Probability

A card is drawn at random from a deck of 52 cards. What is the probability that the drawn card is

1. a queen or a club card
2. a queen or a black card

$1$ card from $52$ cards can be selected in $52$ ways

$\therefore \;$ Number of elements in sample space $S = n\left(S\right) = 52$

Let,

$A$ be the event of drawing a queen card $\;\;\;$ (from $4$ queen cards);

$B$ be the event of drawing a club card $\;\;\;\;\;\;$ (from $13$ club cards);

$C$ be the event of drawing a black card $\;\;\;\;\;$ (from $26$ black cards).

Then, $n\left(A\right) = 4$, $n \left(B\right) = 13$, $n \left(C\right) = 26$

Also, $n \left(A \cap B\right) = 1$, $n \left(A \cap C\right) = 2$

1. $P \left(\text{queen OR club}\right) = P \left(A \text{ OR } B\right) = P \left(A \cup B\right)$
   
   $\because \;$ $A$ and $B$ are not mutually exclusive events, $\left(A \cap B\right) \neq 0$
   
   $\begin{aligned} P \left(A \cup B\right) & = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right) \\\\ & = \dfrac{n \left(A\right)}{n \left(S\right)} + \dfrac{n \left(B\right)}{n \left(S\right)} - \dfrac{n \left(A \cap B\right)}{n \left(S\right)} \\\\ & = \dfrac{4}{52} + \dfrac{13}{52} - \dfrac{1}{52} \\\\ & = \dfrac{4}{13} \end{aligned}$
   
   
2. $P \left(\text{queen OR a black card}\right) = P \left(A \text{ OR } C\right) = P \left(A \cup C\right)$
   
   $\because \;$ $A$ and $C$ are not mutually exclusive events, $\left(A \cap C\right) \neq 0$
   
   $\begin{aligned} P \left(A \cup C\right) & = P \left(A\right) + P \left(C\right) - P \left(A \cap C\right) \\\\ & = \dfrac{n \left(A\right)}{n \left(S\right)} + \dfrac{n \left(C\right)}{n \left(S\right)} - \dfrac{n \left(A \cap C\right)}{n \left(S\right)} \\\\ & = \dfrac{4}{52} + \dfrac{26}{52} - \dfrac{2}{52} \\\\ & = \dfrac{7}{13} \end{aligned}$

Probability

If $A$ and $B$ are mutually exclusive events and $P\left(A\right) = 0.28$, $P \left(B\right)=0.44$, then find

1. $P \left(\overline{A}\right)$
2. $P \left(A \cup B\right)$
3. $P \left(A \cap \overline{B}\right)$
4. $P \left(\overline{A} \cap \overline{B}\right)$

$\because \;$ $A$ and $B$ are mutually exclusive events, $A \cap B = 0$

1. $P \left(\overline{A}\right) = 1 - P \left(A\right) = 1 - 0.28 = 0.72$



2. $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right) = P \left(A\right) + P \left(B\right)$
   
   i.e. $\;$ $P \left(A \cup B\right) = = 0.28 + 0.44 = 0.72 $



3. $P \left(A \cap \overline{B}\right) = P \left(A\right) - P \left(A \cap B\right) = P \left(A\right) = 0.28$



4. $P \left(\overline{A} \cap \overline{B}\right) = P \left(\overline{A \cup B}\right) = 1 - P \left(A \cup B\right) = 1 - 0.72 = 0.28$

Probability

An integer is chosen at random from the first fifty positive integers. What is the probability that the integer chosen is a prime or multiple of $4$.

An integer can be chosen from first fifty positive integers in $50$ ways.

$\therefore \;$ Number of elements in sample space $S = n\left(S\right) = 50$

Let $A$ be the event of selecting an integer which is a prime number OR a multiple of $4$.

There are $15$ prime numbers between $1$ and $50$

$\left(2, \; 3, \; 5, \; 7, \; 11, \; 13, \; 17, \; 19, \; 23, \; 29, \; 31, \; 37, \; 41, \; 43, \; 47\right)$

$\therefore \;$ $1$ prime number can be selected at random from the first fifty positive integers in ${^{15}}{C}_{1} = 15$ ways

There are $12$ numbers which are multiple of $4$ between $1$ and $50$

$\left(4, \; 8, \; 12, \; 16, \; 20, \; 24, \; 28, \; 32, \; 36, \; 40, \; 44, \; 48\right)$

$\therefore \;$ $1$ number which is a multiple of $4$ can be selected from $12$ numbers in ${^{12}}{C}_{1} = 12$ ways

$\therefore \;$ Number of ways in which an integer can be chosen which is a prime number OR a multiple of $4$ $= 15 + 12 = 27$ ways

$\therefore \;$ Number of elements in $A = n\left(A\right) = 15 + 12 = 27$

$\therefore \;$ Probability of $A = P \left(A\right) = \dfrac{n\left(A\right)}{n\left(S\right)} = \dfrac{27}{50}$

Probability

Out of $10$ meritorious students in a school, there are $6$ girls and $4$ boys. A team of $4$ students is selected at random for a quiz program. Find the probability that there are at least $2$ girls.

$4$ students can be selected from $10$ students in

${^{10}}{C}_{4} = \dfrac{10!}{6! \times 4!} = \dfrac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$ ways

$\therefore \;$ Number of elements in sample space $S = n\left(S\right) = 210$

Let $A$ be the event of selecting $4$ students so that there are at least $2$ girls i.e. selecting $2$ girl students OR $3$ girl students OR $4$ girl students

Now,

$2$ girls can be selected from $6$ girls in ${^{6}}{C}_{2} = \dfrac{6!}{4! \times 2!} = \dfrac{6 \times 5}{2 \times 1} = 15$ ways

$\hspace{4cm}$ AND

$2$ boys can be selected from $4$ boys in ${^{4}}{C}_{2} = \dfrac{4!}{2! \times 2!} = \dfrac{4 \times 3}{2 \times 1} = 6$ ways

$\therefore \;$ $2$ girls and $2$ boys can be selected in $15 \times 6 = 90$ ways

$\hspace{4cm}$ OR

$3$ girls can be selected from $6$ girls in ${^{6}}{C}_{3} = \dfrac{6!}{3! \times 3!} = \dfrac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ ways

$\hspace{4cm}$ AND

$1$ boy can be selected from $4$ boys in ${^{4}}{C}_{1} = 4$ ways

$\therefore \;$ $2$ girls and $1$ boy can be selected in $20 \times 4 = 80$ ways

$\hspace{4cm}$ OR

$4$ girls can be selected from $6$ girls in ${^{6}}{C}_{4} = \dfrac{6!}{2! \times 4!} = \dfrac{6 \times 5}{2 \times 1} = 15$ ways

$\therefore \;$ Number of elements in $A = n\left(A\right) = 90 + 80 + 15 = 185$ ways

$\therefore \;$ Probability of event $A = P\left(A\right) = \dfrac{n\left(A\right)}{n\left(S\right)} = \dfrac{185}{210} = \dfrac{37}{42}$

Probability

In a box containing $10$ bulbs, $2$ are defective. What is the probability that among $5$ bulbs chosen at random, none is defective.

$5$ bulbs can be chosen from $10$ bulbs in

${^{10}}{C}_{5} = \dfrac{10!}{5! \times 5!} = \dfrac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$ ways

$\therefore \;$ Number of elements in sample space $S = n\left(S\right) = 252$

$\because \;$ $2$ bulbs are defective, number of bulbs which are not defective $= 8$

Let $A$ be the event of selecting $5$ bulbs which are not defective.

Now, $5$ bulbs can be selected from $8$ non-defective bulbs in

${^{8}}{C}_{5} = \dfrac{8!}{3! \times 5!} = \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$ ways

$\therefore \;$ Number of elements in $A = n\left(A\right) = 56$

$\therefore \;$ Probability of event $A = P\left(A\right) = \dfrac{n\left(A\right)}{n \left(S\right)} = \dfrac{56}{252} = \dfrac{2}{9}$

Probability

A single card is drawn from a pack of $52$ cards. What is the probability that

1. the card is a jack or a king;
2. the card will be $5$ or smaller.

$1$ card from a pack of $52$ cards can be drawn in $52$ ways.

$\therefore \;$ Number of elements in sample space $S = n \left(S\right) = 52$

1. Let $A$ be the event that the selected card is a jack or a king
   
   $1$ jack can be selected from $4$ jacks in ${^{4}}{C}_{1} = 4$ ways
   
   $\hspace{3cm}$ OR
   
   $1$ king can be selected from $4$ kings in ${^{4}}{C}_{1} = 4$ ways
   
   $\therefore \;$ Number of elements in $A = n\left(A\right) = 4 + 4 = 8$
   
   $\therefore \;$ Probability of event $A = P\left(A\right) = \dfrac{n\left(A\right)}{n \left(S\right)} = \dfrac{8}{52} = \dfrac{2}{13}$
   
   
2. Let $B$ be the event that the selected card is $5$ or smaller
   
   There are $16$ such cards $\;\;\;$ (2, 3, 4, 5 of each suite)
   
   $1$ card from $16$ cards can be selected in $16$ ways
   
   $\therefore \;$ Number of elements in $B = n\left(B\right) = 16$
   
   $\therefore \;$ Probability of event $B = P\left(B\right) = \dfrac{n\left(B\right)}{n \left(S\right)} = \dfrac{16}{52} = \dfrac{4}{13}$

Probability

Three coins are tossed once. Find the probability of getting

1. exactly two heads;
2. at least two heads;
3. at most two heads.

When $3$ coins are tossed, number of elements in sample space $S = n\left(S\right) = 2^3 = 8$

1. Let $A$ be the event of getting exactly $2$ heads
   
   Then, $A = \left\{\left(H,H,T\right), \left(H,T,H\right), \left(T, H, H\right)\right\}$
   
   $\therefore \;$ Number of elements in $A = n\left(A\right) = 3$
   
   $\therefore \;$ Probability of event $A = P\left(A\right) = \dfrac{n\left(A\right)}{n \left(S\right)} = \dfrac{3}{8}$
   
   
2. Let $B$ be the event of getting at least $2$ heads
   
   Then, $B = \left\{\left(H,H,T\right), \left(H,T,H\right), \left(T, H, H\right), \left(H,H,H\right)\right\}$
   
   $\therefore \;$ Number of elements in $B = n\left(B\right) = 4$
   
   $\therefore \;$ Probability of event $B = P\left(B\right) = \dfrac{n\left(B\right)}{n \left(S\right)} = \dfrac{4}{8} = \dfrac{1}{2}$
   
   
3. Let $C$ be the event of getting at most $2$ heads
   
   Then, $C = \left\{\left(H,H,T\right), \left(H,T,H\right), \left(T, H, H\right), \left(H,T,T\right), \right.$
   $\hspace{4cm}$ $\left. \left(T,H,T\right), \left(T,T,H\right), \left(T,T,T\right)\right\}$
   
   $\therefore \;$ Number of elements in $C = n\left(C\right) = 7$
   
   $\therefore \;$ Probability of event $C = P\left(C\right) = \dfrac{n\left(C\right)}{n \left(S\right)} = \dfrac{7}{8}$
   
   ALTERNATIVELY
   
   $\begin{aligned} P\left(\text{getting at most 2 heads}\right) & = 1 - P \left(\text{getting 3 heads}\right) \\\\ & = 1 - \dfrac{1}{8} \\\\ & = \dfrac{7}{8} \end{aligned}$

Probability

In a single throw of two dice, find the probability of obtaining

1. a sum of less than 5;
2. a sum of greater than 10;
3. a sum of 9 or 11.

When two dice are thrown, number of elements in sample space $S = n\left(S\right) = 36$

1. Let $A$ be the event of getting a sum of less than $5$
   
   Then, $A = \left\{\left(1,1\right), \left(1,2\right), \left(1,3\right), \left(2,1\right), \left(2,2\right), \left(3,1\right)\right\}$
   
   $\therefore \;$ Number of elements in $A = n\left(A\right) = 6$
   
   $\therefore \;$ Probability of event $A = P\left(A\right) = \dfrac{n\left(A\right)}{n \left(S\right)} = \dfrac{6}{36} = \dfrac{1}{6}$
   
   
2. Let $B$ be the event of getting a sum greater than $10$
   
   Then, $B = \left\{\left(5,6\right), \left(6,5\right), \left(6,6\right)\right\}$
   
   $\therefore \;$ Number of elements in $B = n\left(B\right) = 3$
   
   $\therefore \;$ Probability of event $B = P\left(B\right) = \dfrac{n\left(B\right)}{n \left(S\right)} = \dfrac{3}{36} = \dfrac{1}{12}$
   
   
3. Let $C$ be the event of getting a sum of $9$ or $11$
   
   Then, $C = \left\{\left(3,6\right), \left(4,5\right), \left(5,4\right), \left(5,6\right), \left(6,3\right), \left(6,5\right)\right\}$
   
   $\therefore \;$ Number of elements in $C = n\left(C\right) = 6$
   
   $\therefore \;$ Probability of event $C = P\left(C\right) = \dfrac{n\left(C\right)}{n \left(S\right)} = \dfrac{6}{36} = \dfrac{1}{6}$

Permutations and Combinations

Find the number of strings of $4$ letters that can be formed with the letters of the word $EXAMINATION$?

There are $11$ letters in the word $EXAMINATION$ of which there are $5$ distinct letters (E, X, M, T, O), $2$ A's, $2$ I's and $2$ N's.

All $4$ letters are distinct

$4$ distinct letters can be selected from $8$ distinct letters (E, X, M, T, O, A, I, N) in

${^{8}}{C}_{4} = \dfrac{8!}{4! \times 4!} = \dfrac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}= 70$ ways

Amongst themselves, the selected $4$ letters can be arranged in $4! = 24$ ways

$\therefore \;$ Number of possible arrangements where all $4$ letters are distinct $= 70 \times 24 = 1680$ ways

$1$ pair of letters, $2$ distinct letters

$1$ pair of letters can be selected from the $3$ pair's of letters in ${^{3}}{C}_{1} = 3$ ways

$2$ distinct letters can be selected from the remaining $7$ distinct letters in

${^{7}}{C}_{2} = \dfrac{7!}{5! \times 2!} = \dfrac{7 \times 6}{2 \times 1} = 21$ ways

Amongst themselves, the selected $4$ letters can be arranged in $\dfrac{4!}{2!} = 12$ ways

$\therefore \;$ Number of possible arrangements where $1$ pair of letters and $2$ letters distinct are selected $= 3 \times 21 \times 12 = 756$ ways

$2$ pairs of letters

$2$ pairs of letters can be selected from the $3$ pairs of letters in ${^{3}}{C}_{2} = 3$ ways

Amongst themselves, the selected $4$ letters can be arranged in $\dfrac{4!}{2! \times 2!} = \dfrac{4 \times 3}{2 \times 1} = 6$ ways

$\therefore \;$ Number of possible arrangements where $2$ pairs of letters are selected $= 3 \times 6 = 18$ ways

$\therefore \;$ Total number of possible strings $= 1680 + 756 + 18 = 2454$ ways

Permutations and Combinations

A box contains two white marbles, three black marbles and four red marbles. In how many ways can three marbles be drawn from the box, if at least one black marble is to be included in the draw?

At least $1$ black marble is to be included in the draw - i.e. we can have $1$ OR $2$ OR $3$ black marbles in the draw.

Total number of marbles $= 2 + 3 + 4 = 9$

When $1$ black marble is selected

$1$ black marble can be selected from $3$ black marbles in $3$ ways

Then, $2$ marbles can be selected from the remaining $6$ ($2$ white and $4$ red) marbles in

${^{6}}{C}_{2} = \dfrac{6!}{4! \times 2!} = \dfrac{6 \times 5}{2} = 15$ ways

$\therefore \;$ Number of ways of selecting $3$ marbles when $1$ black marble is selected $= 3 \times 15 = 45$ ways

When $2$ black marbles are selected

$2$ black marbles can be selected from $3$ black marbles in ${^{3}}{C}_{2} = 3$ ways

Then, $1$ marble can be selected from the remaining $6$ ($2$ white and $4$ red) marbles in

${^{6}}{C}_{1} = 6$ ways

$\therefore \;$ Number of ways of selecting $3$ marbles when $2$ black marbles are selected $= 3 \times 6 = 18$ ways

When $3$ black marbles are selected

$3$ black marbles can be selected from $3$ black marbles in $1$ way

$\therefore \;$ Number of ways of selecting $3$ marbles so that at least $1$ black marble is selected $= 45 + 18 + 1 = 64$ ways

Permutations and Combinations

A committee of 7 people is to be formed from 8 men and 4 women. In how many ways can this be done when the committee consists of

1. exactly $3$ women?
2. at least $3$ women?
3. at most $3$ women?

1. Exactly $3$ women
   
   $3$ women can be selected from $4$ women in ${^{4}}{C}_{3} = \dfrac{4!}{1! \times 3!} = 4$ ways
   
   $4$ men can be selected from $8$ men in ${^{8}}{C}_{4} = \dfrac{8!}{4! \times 4!} = \dfrac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$ ways
   
   $\therefore \;$ Number of ways of forming a committee of $7$ when exactly $3$ women are present in the committee are $4 \times 70 = 280$ ways
   
   
2. At least $3$ women
   
   i.e. $\;$ the committee can have $3$ OR $4$ women members
   
   
   1. Number of ways of forming a committee with $3$ women members $= 280$ ways
   
   
   
   2. $4$ women can be selected from $4$ women members in $1$ way.
      
      $3$ men can be selected from $8$ men in ${^{8}}{C}_{3} = \dfrac{8!}{5! \times 3!} = \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$ ways
   
   
   $\therefore \;$ A committee with $4$ women and $3$ men can be formed in $1 \times 56 = 56$ ways
   
   
   $\therefore \;$ Number of ways of forming a committee of $7$ when at least $3$ women are present in the committee are $280 + 56 = 336$ ways
   
   
3. At most $3$ women
   
   i.e. $\;$ the committee can have $3$ OR $2$ OR $1$ OR $0$ women members
   
   
   1. Number of ways of forming a committee with $3$ women members $= 280$ ways
   
   
   
   2. $2$ women can be selected from $4$ women members in ${^{4}}{C}_{2} = \dfrac{4!}{2! \times 2!} = \dfrac{4 \times 3}{2 \times 1} = 6$ ways.
      
      $5$ men can be selected from $8$ men in ${^{8}}{C}_{5} = \dfrac{8!}{3! \times 5!} = \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$ ways
      
      $\therefore \;$ A committee with $2$ women and $5$ men can be formed in $6 \times 56 = 336$ ways
   
   
   
   3. $1$ woman can be selected from $4$ women members in ${^{4}}{C}_{1} = 4$ ways.
      
      $6$ men can be selected from $8$ men in ${^{8}}{C}_{6} = \dfrac{8!}{2! \times 6!} = \dfrac{8 \times 7}{2 \times 1} = 28$ ways
      
      $\therefore \;$ A committee with $1$ woman and $6$ men can be formed in $4 \times 28 = 112$ ways
   
   
   
   4. When $0$ women members are present i.e. all members are men, then the committee can be formed in ${^{8}}{C}_{7} = 8$ ways
   
   
   
   $\therefore \;$ Number of ways of forming a committee of $7$ when at most $3$ women are present in the committee are $280 + 336 + 112 + 8 = 736$ ways

Permutations and Combinations

Determine the number of $5$ card combinations out of a deck of $52$ cards if there is exactly three aces in each combination.

$3$ aces can be selected from $4$ aces in ${^{4}}{C}_{3} = \dfrac{4!}{1! \times 3!} = 4$ ways

The remaining $2$ cards can be selected from the remaining $52 - 4 = 48$ cards in

${^{48}}{C}_{2} = \dfrac{48!}{46! \times 2!} = \dfrac{48 \times 47}{2} = 1128$ ways

$\therefore \;$ Number of $5$ card combinations that can be made from a deck of $52$ cards so that there are exactly $3$ aces in each combination $= 4 \times 1128 = 4512$ combinations

Permutations and Combinations

There are $5$ teachers and $20$ students. Out of them a committee of $2$ teachers and $3$ students is to be formed. Find the number of ways in which this can be done. Further find in how many of these committees

1. a particular teacher is included?
2. a particular student is excluded?

$2$ teachers can be selected from $5$ teachers in

${^{5}}{C}_{2} = \dfrac{5!}{3! \times 2!} = \dfrac{5 \times 4}{2} = 10$ ways

$3$ students can be selected from $20$ students in

${^{20}}{C}_{3} = \dfrac{20!}{17! \times 3!} = \dfrac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$ ways

$\therefore \;$ Number of ways of forming a committee of $2$ teachers and $3$ students from $5$ teachers and $20$ students is $10 \times 1140 = 11400$ ways

1. When a particular teacher is included,
   
   then $1$ teacher is to be selected from $4$ teachers.
   
   This can be done in ${^{4}}{C}_{1} = 4$ ways
   
   $3$ students can be selected from $20$ students in
   
   ${^{20}}{C}_{3} = \dfrac{20!}{17! \times 3!} = \dfrac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$ ways
   
   $\therefore \;$ Number of ways of forming a committee of $2$ teachers and $3$ students from $5$ teachers and $20$ students (when a particular teacher is always included) is $4 \times 1140 = 4560$ ways
   
   
2. When a particular student is always excluded,
   
   then $3$ students can be selected from $19$ students in
   
   ${^{19}}{C}_{3} = \dfrac{19!}{16! \times 3!} = \dfrac{19 \times 18 \times 17}{3 \times 2 \times 1} = 969$ ways
   
   $2$ teachers can be selected from $5$ teachers in
   
   ${^{5}}{C}_{2} = \dfrac{5!}{3! \times 2!} = \dfrac{5 \times 4}{2} = 10$ ways
   
   $\therefore \;$ Number of ways of forming a committee of $2$ teachers and $3$ students from $5$ teachers and $20$ students (when a particular student is always excluded) is $10 \times 969 = 9690$ ways

Permutations and Combinations

How many different selections of $5$ books can be made from $12$ different books if,

1. two particular books are always selected?
2. two particular books are never selected?

1. When $2$ particular books are always selected, then we have to select $3$ books from $10$ books.
   
   $2$ particular books can always be selected from $12$ books in $1$ way
   
   $3$ books can be selected from $10$ books in
   
   ${^{10}}{C}_{3} = \dfrac{10!}{7! \times 3!} = \dfrac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$ ways
   
   $\therefore \;$ Number of ways of selecting $5$ different books from $12$ books when two particular books are always selected $= 1 \times 120 = 120$ ways
   
   
2. When two particular books are never selected, then we have to select $5$ books from $10$ different books.
   
   $5$ books can be selected from $10$ books in
   
   ${^{10}}{C}_{5} = \dfrac{10!}{5! \times 5!} = \dfrac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$ ways

Permutations and Combinations

How many ways a committee of six persons from 10 persons can be chosen along with a chair person and a secretary?

$1$ chair person and $1$ secretary can be selected from $10$ people in

${^{10}}{C}_{2} = \dfrac{10!}{8! \times 2!} = \dfrac{10 \times 9}{2} = 45$ ways.

Out of the remaining $8$ people, $4$ people can be selected in

${^{8}}{C}_{4} = \dfrac{8!}{4! \times 4!}= \dfrac{8 \times 7 \times 6 \times 5 \times 4!}{4! \times 4!} = \dfrac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$ ways

$\therefore \;$ Number of ways a committee of six persons from 10 persons can be chosen along with a chair person and a secretary $= 45 \times 70 = 3150$ ways

Permutations and Combinations

Prove that ${^{2n}}{C}_{n} = \dfrac{2^n \times 1 \times 3 \times \cdots \left(2n - 1\right)}{n!}$

$\begin{aligned} {^{2n}}{C}_{n} & = \dfrac{2n \times \left(2n - 1\right) \times \left(2n - 2\right) \times \left(2n - 3\right) \times \left(2n - 4\right) \times \cdots \times 3 \times 2 \times 1}{\left(2n - n\right)! \times n!} \\\\ & = \dfrac{\left[2n \times \left(2n - 2\right) \times \left(2n - 4\right) \times \cdots \times 2\right] \left[\left(2n - 1\right) \times \left(2n - 3\right) \times \cdots \times 3 \times 1\right]}{n! \times n!} \\\\ & = \dfrac{2^{n} \left[n \times \left(n - 1\right) \times \left(n - 2\right) \times \cdots \times 1\right] \left[\left(2n - 1\right) \times \left(2n - 3\right) \times \cdots \times 3 \times 1\right]}{n! \times n!} \\\\ & = \dfrac{2^{n} \times n! \times \left[\left(2n - 1\right) \times \left(2n - 3\right) \times \cdots \times 3 \times 1\right]}{n! \times n!} \\\\ & = \dfrac{2^{n} \times \left(2n - 1\right) \times \left(2n - 3\right) \times \cdots \times 3 \times 1}{n!} \end{aligned}$

Hence proved.

Permutations and Combinations

Prove that ${^{35}}{C}_{5} + \sum_{r = 0}^{4} {^{\left(39 - r\right)}}{C}_{4} = {^{40}}{C}_{5}$

$\begin{aligned} LHS = {^{35}}{C}_{5} + \sum_{r = 0}^{4} {^{\left(39 - r\right)}}{C}_{4} & = {^{35}}{C}_{5} + {^{39}}{C}_{4} + {^{38}}{C}_{4} + {^{37}}{C}_{4} + {^{36}}{C}_{4} + {^{35}}{C}_{4} \\\\ & = \left({^{35}}{C}_{5} + {^{35}}{C}_{4}\right) + {^{36}}{C}_{4} + {^{37}}{C}_{4} + {^{38}}{C}_{4} + {^{39}}{C}_{4} \\\\ & = \left({^{36}}{C}_{5} + {^{36}}{C}_{4}\right) + {^{37}}{C}_{4} + {^{38}}{C}_{4} + {^{39}}{C}_{4} \\\\ & = \left({^{37}}{C}_{5} + {^{37}}{C}_{4}\right) + {^{38}}{C}_{4} + {^{39}}{C}_{4} \\\\ & = \left({^{38}}{C}_{5} + {^{38}}{C}_{4}\right) + {^{39}}{C}_{4} \\\\ & = {^{39}}{C}_{5} + {^{39}}{C}_{4} \\\\ & = {^{40}}{C}_{5} = RHS \\\\ & \left[\text{Note: } {^{n}}{C}_{r} + {^{n}}{C}_{\left(r - 1\right)} = {^{\left(n + 1\right)}}{C}_{r}\right] \end{aligned}$

Permutations and Combinations

Prove that ${^{15}}{C}_{3} + 2 \times {^{15}}{C}_{4} + {^{15}}{C}_{5} = {^{17}}{C}_{5}$

$\begin{aligned} LHS = {^{15}}{C}_{3} + 2 \times {^{15}}{C}_{4} + {^{15}}{C}_{5} & = {^{15}}{C}_{3} + {^{15}}{C}_{4} + {^{15}}{C}_{4} + {^{15}}{C}_{5} \\\\ & = \left({^{15}}{C}_{4} + {^{15}}{C}_{3}\right) + \left({^{15}}{C}_{5} + {^{15}}{C}_{4}\right) \\\\ & = {^{16}}{C}_{4} + {^{16}}{C}_{5} \;\;\;\;\; \left[\because \; {^{n}}{C}_{r} + {^{n}}{C}_{\left(r - 1\right)} = {^{\left(n + 1\right)}}{C}_{r}\right] \\\\ & = {^{17}}{C}_{5} = RHS \end{aligned}$

Permutations and Combinations

If ${^{\left(n + 2\right)}}{C}_{8} : {^{\left(n - 2\right)}}{P}_{4} = 57 : 16$, find $n$.

Given: $\;$ ${^{\left(n + 2\right)}}{C}_{8} : {^{\left(n - 2\right)}}{P}_{4} = 57 : 16$

i.e. $\;$ $\dfrac{\left(n + 2\right)!}{8! \times \left(n - 6\right)!} \times \dfrac{\left(n - 6\right)!}{\left(n - 2\right)!} = \dfrac{57}{16}$

i.e. $\;$ $\dfrac{\left(n + 2\right) \times \left(n + 1\right) \times \left(n\right) \times \left(n - 1\right)}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2} = \dfrac{57}{16}$

i.e. $\;$ $\left(n + 2\right) \times \left(n + 1\right) \times \left(n\right) \times \left(n - 1\right) = 57 \times 7 \times 6 \times 5 \times 4 \times 3$

i.e. $\;$ $\left(n + 2\right) \times \left(n + 1\right) \times \left(n\right) \times \left(n - 1\right) = 19 \times 3 \times 7 \times 6 \times 5 \times 4 \times 3$

i.e. $\;$ $\left(n + 2\right) \times \left(n + 1\right) \times \left(n\right) \times \left(n - 1\right) = 21 \times 20 \times 19 \times 18$

$\implies$ $n = 19$

Permutations and Combinations

Find $n$ if ${^{n}}{C}_{\left(n - 4\right)} = 70$

Given: $\;$ ${^{n}}{C}_{\left(n - 4\right)} = 70$

i.e. $\;$ $\dfrac{n!}{4! \times \left(n - 4\right)!} = 70$

i.e. $\;$ $\dfrac{n \times \left(n - 1\right) \times \left(n - 2\right) \times \left(n - 3\right) \times \left(n - 4\right)!}{\left(n - 4\right)!} = 4! \times 70$

i.e. $\;$ $n \times \left(n - 1\right) \times \left(n - 2\right) \times \left(n -3\right) = 4 \times 3 \times 2 \times 70$

i.e. $\;$ $n \times \left(n - 1\right) \times \left(n - 2\right) \times \left(n -3\right) = 4 \times 3 \times 2 \times 5 \times 2 \times 7$

i.e. $\;$ $n \times \left(n - 1\right) \times \left(n - 2\right) \times \left(n -3\right) = 8 \times 7 \times 6 \times 5$

$\implies$ $n = 8$

Permutations and Combinations

If ${^{8}}{C}_{r} - {^{7}}{C}_{3} = {^{7}}{C}_{2}$, find $r$

Given: $\;$ ${^{8}}{C}_{r} - {^{7}}{C}_{3} = {^{7}}{C}_{2}$

i.e. $\;$ $\dfrac{8!}{\left(8 - r\right)! \times r!} - \dfrac{7!}{\left(7 - 3\right)! \times 3!} = \dfrac{7!}{\left(7 - 2\right)! \times 2!}$

i.e. $\;$ $\dfrac{8!}{\left(8 - r\right)! \times r!} = \dfrac{7!}{4! \times 3!} + \dfrac{7!}{5! \times 2!}$

i.e. $\;$ $\dfrac{8!}{\left(8 - r\right)! \times r!} = \dfrac{7!}{4! \times 2!} \left(\dfrac{1}{3} + \dfrac{1}{5}\right)$

i.e. $\;$ $\dfrac{8!}{\left(8 - r\right)! \times r!} = \dfrac{8 \times 7!}{4! \times 2! \times 3 \times 5}$

i.e. $\;$ $\dfrac{8!}{\left(8 - r\right)! \times r!} = \dfrac{8!}{5! \times 3!}$

i.e. $\;$ $\left(8 - r\right)! \times r! = 5! \times 3!$

$\implies$ $r! = 3!$ $\implies$ $r = 3$

Permutations and Combinations

If ${^{n}}{C}_{10} = {^{n}}{C}_{12}$, find ${^{23}}{C}_{n}$

Given: $\;$ ${^{n}}{C}_{10} = {^{n}}{C}_{12}$

i.e. $\;$ $\dfrac{n!}{\left(n - 10\right)! \times 10!} = \dfrac{n!}{\left(n - 12\right)! \times 12!}$

i.e. $\;$ $\left(n - 12\right)! \times 12! = \left(n - 10\right)! \times 10!$

i.e. $\;$ $\left(n - 12\right)! \times 12 \times 11 \times 10! = \left(n - 10\right) \times \left(n - 11\right) \times\ \left(n - 12\right)! \times 10!$

i.e. $\;$ $12 \times 11 = \left(n - 10\right) \times \left(n - 11\right)$

$\implies$ $n - 10 = 12$ $\;$ OR $\;$ $n - 11 = 11$

i.e. $\;$ $n = 22$

$\begin{aligned} \therefore \; {^{23}}{C}_{n} & = {^{23}}{C}_{22} \\\\ & = \dfrac{23!}{\left(23 - 22\right)! \times 22!} \\\\ & = \dfrac{23!}{22!} = 23 \end{aligned}$

Permutations and Combinations

In how many ways can a garland of $20$ similar flowers be made?

A garland of $20$ similar flowers can be made in $\left(20 - 1\right)! = 19!$ ways.

This circular arrangement can either be clockwise or anticlockwise.

$\therefore \;$ Number of ways a garland of $20$ similar flowers can be made $= \dfrac{19!}{2}$ ways

Permutations and Combinations

A committee of $11$ members sits at a round table. In how many ways can they be seated if the 'President' and the 'Secretary' choose to sit together.

Consider the President and the Secretary as $1$ person.

Then there are $10$ people to be seated at a round table.

This can be done in $\left(10 - 1\right)! = 9!$ ways.

Amongst themselves, the President and the Secretary can be seated in $2$ ways.

$\therefore \;$ Number of ways a committee of $11$ members sits a round table so that the President and the Secretary sit together $= 2 \times 9!$ ways

Permutations and Combinations

If the letters of the word $GARDEN$ are permuted in all possible ways and the strings thus formed are arranged in the dictionary order, then find the ranks of the words

1. $\; GARDEN$
2. $\; DANGER$

The given word is $GARDEN$.

The dictionary order of the letters of given word is $\; A, \; D, \; E, \; G, \; N, \; R$

1. In the dictionary order, words which begin with $A$ come first.
   
   If the first place is filled with $A$, the remaining 5 letters $\; D, \; E, \; G, \; N, \; R \;$ can be arranged in $5!$ ways.
   
   Proceeding in this manner we have,
   
   $A \; - \; - \; - \; - \; - \; = 5!$ ways
   
   $D \; - \; - \; - \; - \; - \; = 5!$ ways
   
   $E \; - \; - \; - \; - \; - \; = 5!$ ways
   
   $G \; A \; D \; - \; - \; - \; = 3!$ ways
   
   $G \; A \; E \; - \; - \; - \; = 3!$ ways
   
   $G \; A \; N \; - \; - \; - \; = 3!$ ways
   
   $G \; A \; R \; D \; E \; N \; = 1$ way
   
   $\therefore \;$ Rank of the word $GARDEN$ is $= 3 \times 5! + 3 \times 3! + 1 = 3 \times 120 + 3 \times 6 + 1 = 379$



2. Rank of the word $DANGER$
   
   $A \; - \; - \; - \; - \; - \; = 5!$ ways
   
   $D \; A \; E \; - \; - \; - \; = 3!$ ways
   
   $D \; A \; G \; - \; - \; - \; = 3!$ ways
   
   $D \; A \; N \; E \; - \; - \; = 2!$ ways
   
   $D \; A \; N \; G \; E \; R \; = 1$ way
   
   $\therefore \;$ Rank of the word $DANGER$ is $= 5! + 2 \times 3! + 2! + 1 = 120 + 12 + 2 + 1 = 135$

Permutations and Combinations

Each of the digits $\; 1, \; 1, \; 2, \; 3, \; 3, \; 4 \;$ is written on a separate card. The six cards are then laid out in a row to form a 6-digit number.

1. How many distinct 6-digit numbers are there?
2. How many of these 6-digit numbers are even?
3. How many of these 6-digit numbers are divisible by 4?

1. Of the given digits, there are two 1's and two 3's.
   
   $\therefore \;$ Number of distinct 6-digit numbers $= \dfrac{6!}{2! \times 2!} = \dfrac{6 \times 5 \times 4 \times 3}{2} = 180$



2. The unit's place can either be a $2$ or a $4$.
   
   $\therefore \;$ The unit's place can be selected in $2$ ways.
   
   Then the remaining five places can be selected in $\dfrac{5!}{2! \times 2!} = \dfrac{5 \times 4 \times 3}{2} = 30$ ways
   
   $\therefore \;$ Number of 6-digit numbers which are even $= 2 \times 30 = 60$ numbers



3. Required 6-digit numbers which are divisible by 4 are $\; X \; X \; X \; X \; 1 \; 2$ $\;\;$ OR $\;\;$ $\; X \; X \; X \; X \; 2 \; 4$
   
   When numbers are of the form $\; X \; X \; X \; X \; 1 \; 2$:
   
   The unit's place can be selected in $1$ way;
   
   the ten's place can be selected from the two 1's in $2$ ways;
   
   the remaining four places can be selected from the remaining digits $\; 1, \; 3, \; 3, \; 4 \;$ in
   
   $\dfrac{4!}{2!} = \dfrac{4 \times 3 \times 2!}{2!} = 12$ ways
   
   $\therefore \;$ Numbers of the form $\; X \; X \; X \; X \; 1 \; 2$ can be selected in $2 \times 12 = 24$ ways
   
   When numbers are of the form $\; X \; X \; X \; X \; 2 \; 4$:
   
   The unit's and the ten's place can be selected in $1$ way each.
   
   The remaining four places can be selected from the remaining digits $\; 1, \; 1, \; 3, \; 3 \;$ in
   
   $\dfrac{4!}{2! \times 2!} = \dfrac{4 \times 3 \times 2!}{2! \times 2} = 6$ ways
   
   $\therefore \;$ Numbers of the form $\; X \; X \; X \; X \; 2 \; 4$ can be selected in $1 \times 1 \times 6 = 6$ ways
   
   $\therefore \;$ Number of 6-digit numbers which are divisible by 4 $= 24 + 6 = 30$ numbers

Permutations and Combinations

In how many ways can the letters of the word $SUCCESS$ be arranged so that all $Ss$ are together?

The word $SUCCESS$ has $3$ $Ss$.

Consider the three Ss as one S (so that they are always together).

Then, the word $SUCCESS$ has $5$ letters of which there are $2$ $Cs$.

$\therefore \;$ Number of ways these 5 letters can be arranged amongst themselves $= \dfrac{5!}{2!} = \dfrac{5 \times 4 \times 3 \times 2!}{2!} = 60$ ways

$\therefore \;$ Number of ways the letters of the word $SUCCESS$ be arranged so that all $Ss$ are together $= 60$ ways.

Permutations and Combinations

In how many ways $4$ mathematics books, $3$ physics books, $2$ chemistry books and $1$ biology book can be arranged on a shelf so that all books of the same subjects are together.

Consider the $4$ mathematics books as a single book;

the $3$ physics books as a single book;

the $2$ chemistry books as one book.

Then there are $4$ books in all.

$4$ books can be arranged on a shelf in $4! = 4 \times 3 \times 2 \times 1 = 24$ ways

Now, the $4$ mathematics books can be arranged amongst themselves in $4! = 24$ ways

The $3$ physics books can be arranged amongst themselves in $3! = 3 \times 2 \times 1 = 6$ ways

The $2$ chemistry books can be arranged amongst themselves in $2! = 2 \times 1 = 2$ ways

$\therefore \;$ All the given books can be arranged on a shelf so that all books of the same subject are together in $\; 24 \times 24 \times 6 \times 2 = 6912$ ways

Permutations and Combinations

$8$ women and $6$ men are standing in a line.

1. How many arrangements are possible if any individual can stand in any position?


2. In how many arrangements will all $6$ men be standing next to one another?


3. In how many arrangements will no two men be standing next to one another?

1. There are $8$ women and $6$ men i.e. $14$ people.
   
   $\therefore \;$ Number of arrangements possible if any individual can stand in any position $= 14!$


2. Consider the $6$ men as ONE group.
   
   Then $8$ women plus the $1$ group of 6 men can stand in a line in $9!$ ways.
   
   Now, the $6$ men can, amongst themselves, stand in a line in $6!$ ways.
   
   $\therefore \;$ Number of ways in which all $6$ men will be standing next to one another $= 9! \times 6!$ ways


3. $W \;\;\; M \;\;\; W \;\;\; M \;\;\; W \;\;\; M \;\;\; W \;\;\; M \;\;\; W \;\;\; M \;\;\; W \;\;\; M \;\;\; W \;\;\; W \;\;\; W$
   
   When no two men are standing next to one another, there are $9$ places which can be occupied by the $6$ men.
   
   $9$ places can be occupied by $6$ men in ${^{9}}{P}_{6}$ ways.
   
   The $8$ women can be arranged amongst themselves in $8!$ ways.
   
   $\therefore \;$ Number of ways in which 8 women and 6 men can be arranged so that no two men are standing next to one another $= 8! \times {^{9}}{P}_{6}$ ways

Permutations and Combinations

How many words can be formed from the letters of the word $\; ARTICLE \;$ so that vowels occupy even places?

The word $\; ARTICLE \;$ has 7 letters with 3 vowels $\; A, \; I, \; E$

There are $3$ even places which the 3 vowels can occupy.

This can be done in $3! = 3 \times 2 \times 1 = 6$ ways

We are left with 4 consonants $\; R, \; T, \; C, \; L$ to occupy the remaining $4$ odd places.

This can be done in $4! = 4 \times 3 \times 2 \times 1 = 24$ ways

$\therefore \;$ Number of words that can be formed from the letters of the word $\; ARTICLE \;$ so that vowels occupy even places $= 6 \times 24 = 144$

Permutations and Combinations

How many 4-digit numbers are there, when a digit may be repeated any number of times?

The available digits are: $\; 0, \; 1, \; 2, \; 3, \; 4, \; 5, \; 6, \; 7, \; 8, \; 9$

$\therefore \;$ Number of available digits $= 10$

The thousand's place can be selected from the digits $\; 1, \; 2, \cdots , 9 \;$ in $\; 9$ ways

The hundred's place can be selected from the digits $\; 0, \; 1, \; 2, \cdots, 9 \;$ in $\; 10$ ways

The ten's place can be selected from the digits $\; 0, \; 1, \; 2, \cdots, 9 \;$ in $\; 10$ ways

The unit's place can be selected from the digits $\; 0, \; 1, \; 2, \cdots, 9 \;$ in $\; 10$ ways

$\therefore \;$ Number of possible 4-digit numbers $= 9 \times 10 \times 10 \times 10 = 9000$

Permutations and Combinations

How many different words can be formed (with or without meaning) with the letters of the word $ALGEBRA$? How many of these begin with $L$ and end with $R$?

The word $\;$ $ALGEBRA$ $\;$ has $7$ letters of which two are $A's$ and the rest are different.

$\therefore \;$ Number of words which can be formed with the letters of the word $ALGEBRA$ are

$= \dfrac{7!}{2!} = \dfrac{7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2!} = 2520$

When the words begin with $L$ and end with $R$, the first and the seventh place can be selected in $1$ way each.

The remaining five places can be selected from the letters $A, \; G, \; E, \; B, \; A$ in

$\dfrac{5!}{2!} = \dfrac{5 \times 4 \times 3 \times 2!}{2!} = 60$ ways

$\therefore \;$ Number of words which begin with $L$ and end with $R$ $= 1 \times 60 \times 1 = 60$ words

Permutations and Combinations

Find the sum of all the numbers that can be formed with the digits $2, \; 3, \; 4, \; 5$ taken all at a time.

When $5$ is in the unit's place, the remaining 3 digits ($2, \; 3, \; 4$) can be arranged amongst themselves in $3! = 6$ ways.

i.e. $\;$ $6$ numbers are possible with 5 in the unit's place.

$\therefore \;$ Sum of unit's place $= 6 \times \left(5 \times 1\right) = 30$

When $5$ is in the ten's place, the remaining 3 digits ($2, \; 3, \; 4$) can be arranged amongst themselves in $3! = 6$ ways.

i.e. $\;$ $6$ numbers are possible with 5 in the ten's place.

$\therefore \;$ Sum of ten's place $= 6 \times \left(5 \times 10\right) = 300$

When $5$ is in the hundred's place, the remaining 3 digits ($2, \; 3, \; 4$) can be arranged amongst themselves in $3! = 6$ ways.

i.e. $\;$ $6$ numbers are possible with 5 in the hundred's place.

$\therefore \;$ Sum of hundred's place $= 6 \times \left(5 \times 100\right) = 3000$

When $5$ is in the thousand's place, the remaining 3 digits ($2, \; 3, \; 4$) can be arranged amongst themselves in $3! = 6$ ways.

i.e. $\;$ $6$ numbers are possible with 5 in the thousand's place.

$\therefore \;$ Sum of thousand's place $= 6 \times \left(5 \times 1000\right) = 30000$

$\therefore \;$ Sum of all possible numbers with 5 in the unit's, ten's, hundred's and thousand's place is

$= 30 + 300 + 3000 + 30000 = 33330$

Similarly, sum of all possible numbers with 4 in the unit's, ten's, hundred's and thousand's place is

$= 6 \times \left(4 \times 1\right) + 6 \times \left(4 \times 10\right) + 6 \times \left(4 \times 100\right) + 6 \times \left(4 \times 1000\right)$

$= 24 + 240 + 2400 + 24000 = 26664$

Sum of all possible numbers with 3 in the unit's, ten's, hundred's and thousand's place is

$= 6 \times \left(3 \times 1\right) + 6 \times \left(3 \times 10\right) + 6 \times \left(3 \times 100\right) + 6 \times \left(3 \times 1000\right)$

$= 18 + 180 + 1800 + 18000 = 19998$

Sum of all possible numbers with 2 in the unit's, ten's, hundred's and thousand's place is

$= 6 \times \left(2 \times 1\right) + 6 \times \left(2 \times 10\right) + 6 \times \left(2 \times 100\right) + 6 \times \left(2 \times 1000\right)$

$= 12 + 120 + 1200 + 12000 = 13332$

$\therefore \;$ Sum of all the numbers that can be formed with the digits $2, \; 3, \; 4, \; 5$ taken all at a time

$= 33330 + 26664 + 19998 + 13332 = 93324$

Permutations and Combinations

How many 3-digit numbers are there, with distinct digits, with each digit odd?

Since each digit of the 3-digit number is odd, the available digits are $1, \; 3, \; 5, \; 7, \; 9$

i.e. $\;$ Number of digits available $= 5$ digits

A 3-digit number can be selected from the 5 digits in

${^{5}}{P}_{3} = \dfrac{5!}{\left(5 - 3\right)!} = \dfrac{5!}{2!} = \dfrac{5 \times 4 \times 3 \times 2!}{2!} = 60$ ways

$\therefore \;$ Number of possible 3-digit numbers, with distinct odd digits is $60$.

Permutations and Combinations

Three men have 4 coats, 5 waistcoats and 6 caps. In how many ways can they wear them?

$4$ Coats can be selected by $3$ men in $\;$ ${^{4}}{P}_{3} = 4! = 24$ ways

$5$ Waistcoats can be selected by $3$ men in

${^{5}}{P}_{3} = \dfrac{5!}{\left(5 - 3\right)!} = \dfrac{5 \times 4 \times 3 \times 2! }{2!} = 60$ ways

$6$ Waistcoats can be selected by $3$ men in

${^{6}}{P}_{3} = \dfrac{6!}{\left(6 - 3\right)!} = \dfrac{6 \times 5 \times 4 \times 3! }{3!} = 120$ ways

$\therefore \;$ Total number of ways of selecting $4$ coats, $5$ waistcoats and $6$ caps by $3$ men

$= 24 \times 60 \times 120 = 172800$ ways

Permutations and Combinations

Prove that ${^{n}}{P}_{r} = {^{\left(n - 1\right)}}{P}_{r} + r \cdot {^{\left(n - 1\right)}}{P}_{\left(r - 1\right)}$

By definition,

${^{n}}{P}_{r} = \dfrac{n!}{\left(n - r\right)!}$ $\;\;\; \cdots \; (1)$

${^{\left(n - 1\right)}}{P}_{r} = \dfrac{\left(n - 1\right)!}{\left(n - r - 1\right)!}$ $\;\;\; \cdots \; (2)$

${^{\left(n - 1\right)}}{P}_{\left(r - 1\right)} = \dfrac{\left(n - 1\right)!}{\left(n - r\right)!}$ $\;\;\; \cdots \; (3)$

$\therefore \;$ We have from equations $(2)$ and $(3)$,

$\begin{aligned} {^{\left(n - 1\right)}}{P}_{r} + r \cdot {^{\left(n - 1\right)}}{P}_{\left(r - 1\right)} & = \dfrac{\left(n - 1\right)!}{\left(n - r - 1\right)!} + r \cdot \dfrac{\left(n - 1\right)!}{\left(n - r\right)!} \\\\ & = \dfrac{\left(n - 1\right)!}{\left(n - r - 1\right)!} + \dfrac{r \cdot \left(n - 1\right)!}{\left(n - r\right) \left(n - r - 1\right)!} \\\\ & = \dfrac{\left(n - 1\right)!}{\left(n - r - 1\right)!} \left[1 + \dfrac{r}{n - r}\right] \\\\ & = \dfrac{\left(n - 1\right)!}{\left(n - r - 1\right)!} \left[\dfrac{n}{n - r}\right] \\\\ & = \dfrac{n!}{\left(n - r\right)!} \\\\ & = {^{n}}{P}_{r} \;\;\;\; \left[\text{from equation (1)}\right] \end{aligned}$

Hence proved.

Permutations and Combinations

If $\;$ ${^{56}}{}{P}_{\left(r + 6\right)} : {^{54}}{}{P}_{\left(r + 3\right)} = 30800 : 1$, find $r$.

Given: $\;$ ${^{56}}{}{P}_{\left(r + 6\right)} : {^{54}}{}{P}_{\left(r + 3\right)} = 30800 : 1$

i.e. $\;$ $\dfrac{56!}{\left(56 - r - 6\right)!} : \dfrac{54!}{\left(54 - r - 3\right)!} = 30800 : 1$

i.e. $\;$ $\dfrac{56!}{\left(50 - r\right)!} \times \dfrac{\left(51 - r\right)!}{54!} = \dfrac{30800}{1}$

i.e. $\;$ $\dfrac{56 \times 55 \times 54!}{\left(50 - r\right)!} \times \dfrac{\left(51 - r\right) \left(50 - r\right)!}{54!} = 30800$

i.e. $\;$ $56 \times 55 \times \left(51 - r\right) = 30800$

i.e. $\;$ $51 - r = \dfrac{30800}{56 \times 55} = 10$

i.e. $\;$ $r = 51 - 10 = 41$

Permutations and Combinations

How many three-letter words can be formed using the letters $\;$ $a, \; b, \; c, \; d, \; e$ $\;$ if:

1. repetition is allowed;
2. repetition is not allowed?

Given: 5 letters $\; a, \; b, \; c, \; d, \; e$

1. Repetition is allowed
   The first letter can be selected in $5$ ways.
   
   The second letter can be selected in $5$ ways.
   
   The third letter can be selected in $5$ ways.
   
   $\therefore \;$ Total number of ways of forming three-letter words using the given letters if repetition is allowed is $= 5 \times 5 \times 5 = 125$ ways.
   
   
2. Repetition is not allowed
   The first letter can be selected in $5$ ways.
   
   The second letter can be selected in $4$ ways.
   
   The third letter can be selected in $3$ ways.
   
   $\therefore \;$ Total number of ways of forming three-letter words using the given letters if repetition is allowed is $= 5 \times 4 \times 3 = 60$ ways.

Permutations and Combinations

How many odd numbers less than $1000$ can be formed by using the digits $0, \; 3, \; 5, \; 7$ when repetition of digits is not allowed?

The required numbers are 3 digit OR 2 digit OR 1 digit odd numbers.

Case 1: 3-digit odd numbers

The unit's place can be selected from the digits $3, \; 5, \;$ and $7$ in $3$ ways.

The ten's place can either be a 0 or can be selected from the remaining odd digits.

If the ten's place is 0, it can be selected in $1$ way.

OR, the ten's place can be selected from remaining two odd digits in $2$ ways.

When the ten's place is 0, then the hundred's place can be selected from the remaining two odd digits in $2$ ways.

OR, when the ten's place is an odd digit, then the hundred's place can be selected from the remaining odd digit in $1$ way.

$\therefore \;$ Number of ways of forming a 3-digit odd number less than $1000$ using the given digits is

$= 3 \times 1 \times 2 + 3 \times 2 \times 1 = 12$ ways

Case 2: 2-digit odd numbers

The unit's place can be selected from the digits $3, \; 5, \;$ and $7$ in $3$ ways.

The ten's place can be selected from the remaining two odd digits in $2$ ways.

$\therefore \;$ Number of ways of forming a 3-digit odd number less than $1000$ using the given digits is $= 3 \times 2 = 6$ ways

Case 3: 1-digit odd numbers

Number of ways of forming 1-digit odd numbers from the given digits is $3$ ways

$\therefore \;$ Total number ways of forming odd numbers less than $1000$ by using the given digits is $12 + 6 + 3 = 21$ ways

i.e. $\;$ $21$ odd numbers can be formed from the given digits which are less than 1000.

Permutations and Combinations

How many different numbers of six digits can be formed from the digits $2, \; 3, \; 0, \; 7, \; 9, \; 5$ when repetition of digits is not allowed?

The given digits are $2, \; 3, \; 0, \; 7, \; 9, \; 5$

$\because$ $\;$ The lakh's place cannot be a $0$, it can be selected from the given digits in $5$ ways.

Now, the ten-thousand's place can be selected from the remaining 5 digits (since repetition of digits is not allowed) in $5$ ways.

The thousand's place can be selected from the remaining 4 digits in $4$ ways.

The hundred's place can be selected from the remaining 3 digits in $3$ ways.

The ten's place can be selected from the remaining 2 digits in $2$ ways.

The unit's place can be selected from the remaining 1 digit in $1$ ways.

$\therefore$ $\;$ Total number of ways of forming a six digit number using the given digits (without repetition) is $5 \times 5 \times 4 \times 3 \times 2 \times 1 = 600$ ways

Permutations and Combinations

How many numbers are there between $500$ and $1000$ which have exactly one of their digits as $8$?

We have $3$ digit numbers between $500$ and $1000$.

Case 1: $\;$ If the unit's place is $8$, then

the unit's place can be selected in $1$ way;

the ten's place can be selected from the digits $0, \; 1, \; 2, \; 3, \; 4, \; 5, \; 6, \; 7, \; 9$ $\;$ i.e. $\;$ in $9$ ways;

the hundred's place can be selected from the digits $5, \; 6, \; 7, \; 9$ $\;$ i.e. $\;$ in $4$ ways.

$\therefore$ $\;$ Number of ways of having a 3 digit number with 8 in the unit's place $= 1 \times 9 \times 4 = 36$ ways

Case 2: $\;$ If the ten's place is $8$, then

the unit's place can be selected from the digits $0, \; 1, \; 2, \; 3, \; 4, \; 5, \; 6, \; 7, \; 9$ $\;$ i.e. $\;$ in $9$ ways;

the ten's place can be selected in $1$ way;

the hundred's place can be selected from the digits $5, \; 6, \; 7, \; 9$ $\;$ i.e. $\;$ in $4$ ways.

$\therefore$ $\;$ Number of ways of having a 3 digit number with 8 in the ten's place $= 9 \times 1 \times 4 = 36$ ways

Case 3: $\;$ If the hundred's place is $8$, then

the unit's place can be selected from the digits $0, \; 1, \; 2, \; 3, \; 4, \; 5, \; 6, \; 7, \; 9$ $\;$ i.e. $\;$ in $9$ ways;

the ten's place can be selected from the digits $0, \; 1, \; 2, \; 3, \; 4, \; 5, \; 6, \; 7, \; 9$ $\;$ i.e. $\;$ in $9$ ways;

the hundred's place can be selected in $1$ way;

$\therefore$ $\;$ Number of ways of having a 3 digit number with 8 in the ten's place $= 9 \times 9 \times 1 = 81$ ways

$\therefore$ $\;$ Total number of numbers between $500$ and $1000$ which have exactly one of their digits as $8$

$=$ Case 1 $\;$ OR $\;$ Case 2 $\;$ OR $\;$ Case 3

$= 36 + 36 + 81 = 153$ numbers

Permutations and Combinations

There are 6 multiple choice questions in an examination. How many sequences of answers are possible, if the first three questions have 4 choices each and the next three have 2 choices each?

The first three questions have 4 choices each.

$\therefore$ $\;$ Each of the first three questions can be answered in $4$ ways each.

$\therefore$ $\;$ The first three questions can be answered in a total of $4 \times 4 \times 4 = 64$ ways.

The next three questions have 2 choices each.

$\therefore$ $\;$ Each of these questions can be answered in $2$ ways each.

$\therefore$ $\;$ The next three questions can be answered in a total of $2 \times 2 \times 2 = 8$ ways.

$\therefore$ $\;$ The total number of ways the 6 multiple choice questions can be answered $= 64 \times 8 = 512$ ways.

Complex Numbers

Solve $\;$ $x^4 + 4 = 0$

Given: $\;$ $x^4 + 4 = 0$

i.e. $\;$ $\left(x^2 + 2i\right) \left(x^2 - 2i\right)= 0$

$\implies$ $\left(x^2 + 2i\right) = 0$ $\;$ OR $\;$ $\left(x^2 - 2i\right) = 0$

Now, $x^2 - 2i = 0$ $\implies$ $x = \left(2i\right)^{\frac{1}{2}}$

Let $2i = r \left(\cos \theta_1 + i \sin \theta_1\right)$

$\implies$ $r \cos \theta_1 = 0$, $\;\;$ $r \sin \theta_1 =2 $

$\therefore$ $\;$ $r = \sqrt{\left(0\right)^2 + \left(2\right)^2} = 2$

$\therefore$ $\;$ $\cos \theta_1 = 0$, $\;$ $\sin \theta_1 = 1$ $\implies$ $\theta_1 = \dfrac{\pi}{2}$

$\therefore$ $\;$ $2i = 2 \left[\cos \left(\dfrac{\pi}{2}\right) + i \sin \left(\dfrac{\pi}{2}\right)\right]$

$\begin{aligned} \therefore \; x = \left(2 i\right)^{\frac{1}{2}} & = 2^{\frac{1}{2}} \left[\cos \left(\dfrac{\pi}{2}\right) + i \sin \left(\dfrac{\pi}{2}\right)\right]^{\frac{1}{2}} \\\\ & = \sqrt{2} \left[\cos \left(2 k \pi + \dfrac{\pi}{2}\right) + i \sin \left(2 k \pi + \dfrac{\pi}{2}\right)\right]^{\frac{1}{2}} \\\\ & = \sqrt{2} \left[\cos \left(4 k + 1\right) \dfrac{\pi}{4} + i \sin \left(4 k + 1\right) \dfrac{\pi}{4}\right] \;\;\;\; k = 0,1 \\\\ & = \sqrt{2} \left(\cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{4}\right), \; \sqrt{2} \left(\cos \dfrac{5 \pi}{4} + i \sin \dfrac{5 \pi}{4}\right) \end{aligned}$

Now, $x^2 + 2i = 0$ $\implies$ $x = \left(-2i\right)^{\frac{1}{2}}$

Let $-2i = R \left(\cos \theta_2 + i \sin \theta_2\right)$

$\implies$ $R \cos \theta_2 = 0$, $\;\;$ $R \sin \theta_2 = - 2 $

$\therefore$ $\;$ $R = \sqrt{\left(0\right)^2 + \left(-2\right)^2} = 2$

$\therefore$ $\;$ $\cos \theta_2 = 0$, $\;$ $\sin \theta_2 = - 1$ $\implies$ $\theta_2 = 2 \pi - \dfrac{\pi}{2} = \dfrac{3 \pi}{2}$

$\begin{aligned} \therefore \; x = \left(-2 i\right)^{\frac{1}{2}} & = 2^{\frac{1}{2}} \left[\cos \left(\dfrac{3\pi}{2}\right) + i \sin \left(\dfrac{3\pi}{2}\right)\right]^{\frac{1}{2}} \\\\ & = \sqrt{2} \left[\cos \left(2 k \pi + \dfrac{3\pi}{2}\right) + i \sin \left(2 k \pi + \dfrac{3\pi}{2}\right)\right]^{\frac{1}{2}} \\\\ & = \sqrt{2} \left[\cos \left(4 k + 3\right) \dfrac{\pi}{4} + i \sin \left(4 k + 3\right) \dfrac{\pi}{4}\right] \;\;\;\; k = 0,1 \\\\ & = \sqrt{2} \left(\cos \dfrac{3\pi}{4} + i \sin \dfrac{3\pi}{4}\right), \; \sqrt{2} \left(\cos \dfrac{7 \pi}{4} + i \sin \dfrac{7 \pi}{4}\right) \end{aligned}$

$\therefore$ $\;$ $x = \sqrt{2} \; cis \left(\dfrac{\pi}{4}\right), \; \sqrt{2} \; cis \left(\dfrac{3\pi}{4}\right), \; \sqrt{2} \; cis \left(\dfrac{5\pi}{4}\right), \; \sqrt{2} \; cis \left(\dfrac{7\pi}{4}\right)$

Complex Numbers

Prove that if $\omega^3 = 1$, then $\dfrac{1}{1 + 2 \omega} - \dfrac{1}{1 + \omega} + \dfrac{1}{2 + \omega} = 0$

Given: $\;$ $\omega^3 = 1$

$\implies$ $\omega$ is the cube root of unity.

Then, $1 + \omega + \omega^2 = 0$

Now,

$\begin{aligned} \dfrac{1}{1 + 2 \omega} - \dfrac{1}{1 + \omega} & = \dfrac{1 + \omega - 1 - 2 \omega}{\left(1 + 2 \omega\right) \left(1 + \omega\right)} \\\\ & = \dfrac{- \omega}{1 + 3 \omega + 2 \omega^2} \\\\ & = \dfrac{-\omega}{\left(1 + \omega + \omega^2\right) + 2 \omega + \omega^2} \\\\ & = \dfrac{- \omega}{2 \omega + \omega^2} \\\\ & = \dfrac{-1}{2 + \omega} \end{aligned}$

$\begin{aligned} \therefore \; \dfrac{1}{1 + 2 \omega} - \dfrac{1}{1 + \omega} + \dfrac{1}{2 + \omega} & = \dfrac{-1}{2 + \omega} + \dfrac{1}{2 + \omega} \\\\ & = 0 \end{aligned}$

Hence proved.

Complex Numbers

Prove that if $\omega^3 = 1$, then $\left(\dfrac{-1 + i \sqrt{3}}{2}\right)^5 + \left(\dfrac{-1 - i \sqrt{3}}{2}\right)^5 = -1$

Given: $\;$ $\omega^3 = 1$

$\implies$ $\omega$ is the cube root of unity.

Then, $\omega = \dfrac{-1 + i \sqrt{3}}{2}$ $\;$ and $\;$ $\omega^2 = \dfrac{-1 - i \sqrt{3}}{2}$

Now,

$\begin{aligned} \left(\dfrac{-1 + i \sqrt{3}}{2}\right)^5 + \left(\dfrac{-1 - i \sqrt{3}}{2}\right)^5 & = \left(\omega\right)^5 + \left(\omega^2\right)^5 \\\\ & = \omega^3 \times \omega^2 + \left(\omega^3\right)^3 \times \omega \\\\ & = \omega^2 + \omega \;\;\; \left[\because \; \omega^3 = 1\right] \\\\ & = -1 \;\;\; \left[\because \; 1 + \omega + \omega^2 = 0\right] \end{aligned}$

Hence proved.

Complex Numbers

If $\;$ $x = a + b$, $\;$ $y = a \omega + b \omega^2$ $\;$ and $\;$ $z = a \omega^2 + b \omega$, $\;$ show that $\;$ $x^3 + y^3 + z^3 = 3 \left(a^3 + b^3\right)$ $\;$ where $\omega$ is the complex cube root of unity.

$x= a + b$

$\therefore$ $\;$ $x^3 = \left(a + b\right)^3 = a^3 + b^3 + 3 a^2 b + 3 a b^2$ $\;\;\; \cdots \; (1)$

$y = a\omega + b \omega^2$

$\begin{aligned} \therefore \; y^3 & = \left(a \omega + b \omega^2\right)^3 \\\\ & = a^3 \omega^3 + b^3 \left(\omega^2\right)^3 + 3 a^2 b \omega^4 + 3 a b^2 \omega^5 \\\\ & = a^3 \omega^3 + b^3 \left(\omega^3\right)^2 + 3 a^2 b \omega^3 \times \omega + 3 a b^2 \omega^3 \times \omega^2 \\\\ & = a^3 + b^3 + 3 a^2 b \omega + 3 ab^2 \omega^2 \;\;\; \cdots \; (2) \;\;\; [\text{Note: } \omega^3 = 1] \end{aligned}$

$z = a \omega^2 + b \omega$

$\begin{aligned} \therefore \; z^3 & = \left(a \omega^2 + b \omega\right)^3 \\\\ & = a^3 \left(\omega^2\right)^3 + b^3 \omega^3 + 3 a^2 b \omega^5 + 3 a b^2 \omega^4 \\\\ & = a^3 \left(\omega^3\right)^2 + b^3 \omega^3 + 3 a^2 b \omega^3 \times \omega^2 + 3 a b^2 \omega^3 \times \omega \\\\ & = a^3 + b^3 + 3 a^2 b \omega^2 + 3 a b^2 \omega \;\;\; \cdots \; (3) \end{aligned}$

$\therefore$ $\;$ We have from equations $(1)$, $(2)$ and $(3)$,

$\begin{aligned} x^3 + y^3 + z^3 & = a^3 + b^3 + 3 a^2 b + 3 ab^2 \\ & \hspace{1cm} a^3 + b^3 + 3 a^2 b \omega + 3 ab^2 \omega^2 \\ & \hspace{2cm} a^3 + b^3 + 3 a^2 b \omega^2 + 3 ab^2 \omega \\\\ & = 3 a^3 + 3 b^3 + 3 a^2 b \left(1 + \omega + \omega^2\right) + 3 ab^2 \left(1 + \omega^2 + \omega\right) \\\\ & = 3 \left(a^3 + b^3\right) \;\;\; [\because \; 1 + \omega + \omega^2 = 0] \end{aligned}$

Hence proved.

Complex Numbers

Find the value of $\;$ $\left(- \sqrt{3} - i\right)^{\frac{2}{3}}$

Let $\left(- \sqrt{3} - i\right) = r \left(\cos \theta + i \sin \theta\right)$

Then, $\;$ $r \cos \theta = - \sqrt{3}$; $\;\;$ $r \sin \theta = -1$

$\therefore$ $\;$ $r = \sqrt{\left(- \sqrt{3}\right)^2 + \left(-1\right)^2} = 2$

Now, $\;$ $\cos \theta = - \dfrac{\sqrt{3}}{2}$; $\;$ $\sin \theta = - \dfrac{1}{2}$ $\implies$ $\theta = -\pi + \dfrac{\pi}{6} = \dfrac{- 5 \pi}{6}$

$\begin{aligned} \therefore \; \left(- \sqrt{3} - i\right)^{\frac{2}{3}} & = 2^{\frac{2}{3}} \left[\cos \left(\dfrac{-5\pi}{6}\right) + i \sin \left(\dfrac{-5 \pi}{6}\right)\right]^{\frac{2}{3}} \\\\ & = 2 ^{\frac{2}{3}} \left\{\left[\cos \left(\dfrac{-5 \pi}{6}\right) + i \sin \left(\dfrac{-5 \pi}{6}\right)\right]^2\right\}^{\frac{1}{3}} \\\\ & = 2^{\frac{2}{3}} \left[\cos \left(\dfrac{-5 \pi}{3}\right) + i \sin \left(\dfrac{-5 \pi}{3}\right)\right]^{\frac{1}{3}} \\\\ & = 2^{\frac{2}{3}} \left[\cos \left(2 k \pi - \dfrac{5 \pi}{3}\right) + i \sin \left(2 k \pi - \dfrac{5 \pi}{3}\right)\right]^{\frac{1}{3}} \\\\ & = 2^{\frac{2}{3}} \left\{\cos \left[\left(6k - 5\right) \dfrac{\pi}{9}\right] + i \sin \left[\left(6k - 5\right) \dfrac{\pi}{9}\right] \right\} \;\;\; where \; k = 0, 1, 2 \end{aligned}$

$\therefore$ $\;$ The values of $\;$ $\left(- \sqrt{3} - i\right)^{\frac{2}{3}}$ $\;$ are

$2^{\frac{2}{3}} \left[\cos \left(\dfrac{-5 \pi}{9}\right) + i \sin \left(\dfrac{-5 \pi}{9}\right)\right]$, $\;$ $2^{\frac{2}{3}} \left[\cos \left(\dfrac{\pi}{9}\right) + i \sin \left(\dfrac{\pi}{9}\right)\right]$, $\;$ $2^{\frac{2}{3}} \left[\cos \left(\dfrac{7\pi}{9}\right) + i \sin \left(\dfrac{7\pi}{9}\right)\right]$

Complex Numbers

If $\;$ $a = \cos 2 \alpha + i \sin 2 \alpha$, $\;$ $b = \cos 2 \beta + i \sin 2 \beta$ $\;$ and $\;$ $c = \cos 2 \gamma + i \sin 2 \gamma$, $\;$ prove that $\;$ $\dfrac{a^2 b^2 + c^2}{abc} = 2 \cos 2 \left(\alpha + \beta - \gamma\right)$

Given: $\;$ $a = \cos 2 \alpha + i \sin 2 \alpha$, $\;$ $b = \cos 2 \beta + i \sin 2 \beta$, $\;$ $c = \cos 2 \gamma + i \sin 2 \gamma$

$\therefore$ $\;$ $a^2 = \left(\cos 2 \alpha + i \sin 2 \alpha\right)^2 = \cos 4 \alpha + i \sin 4 \alpha$

$b^2 = \left(\cos 2 \beta + i \sin 2 \beta\right)^2 = \cos 4 \beta + i \sin 4 \beta$

$c^2 = \left(\cos 2 \gamma + i \sin 2 \gamma\right)^2 = \cos 4 \gamma + i \sin 4 \gamma$

Now,

$\begin{aligned} a^2 b^2 & = \left(\cos 4 \alpha + i \sin 4 \alpha\right) \left(\cos 4 \beta + i \sin 4 \beta\right) \\\\ & = \left(\cos 4 \alpha \; \cos 4 \beta - \sin 4 \alpha \; \sin 4 \beta\right) + i \left(\sin 4 \alpha \; \cos 4 \beta + \cos 4 \alpha \; \sin 4 \beta\right) \\\\ & = \cos \left(4 \alpha + 4 \beta\right) + i \sin \left(4 \alpha + 4 \beta\right) \end{aligned}$

$\begin{aligned} \therefore \; a^2 b^2 + c^2 & = \left[\cos \left(4 \alpha + 4 \beta\right) + i \sin \left(4 \alpha + 4 \beta\right)\right] + \left[\cos 4 \gamma + i \sin 4 \gamma\right] \\\\ & = \left[\cos \left(4 \alpha + 4 \beta\right) + \cos 4 \gamma\right] + i \left[\sin \left(4 \alpha + 4 \beta\right) + \sin 4 \gamma\right] \\\\ & = 2 \cos \left(\dfrac{4 \alpha + 4 \beta + 4 \gamma}{2}\right) \cos \left(\dfrac{4 \alpha + 4 \beta - 4 \gamma}{2}\right) \\ & \hspace{1cm} + 2 i \sin \left(\dfrac{4 \alpha + 4 \beta + 4 \gamma}{2}\right) \cos \left(\dfrac{4 \alpha + 4 \beta - 4 \gamma}{2}\right) \\\\ & = 2 \cos \left(2 \alpha + 2 \beta - 2 \gamma\right)\left[\cos \left(2 \alpha + 2 \beta + 2 \gamma\right) + i \sin \left(2 \alpha + 2 \beta + 2 \gamma\right)\right] \end{aligned}$

$\begin{aligned} abc & = \left(\cos 2 \alpha + i \sin 2 \alpha\right) \left(\cos 2 \beta + i \sin 2 \beta\right) \left(\cos 2 \gamma + i \sin 2 \gamma\right) \\\\ & = \left[\left(\cos 2 \alpha \; \cos 2 \beta - \sin 2 \alpha \; \sin 2 \beta \right) + i \left(\sin 2 \alpha \; \cos 2 \beta + \cos 2 \alpha \; \sin 2 \beta \right)\right] \\ & \hspace{9cm} \times \left(\cos 2 \gamma + i \sin 2 \gamma\right) \\\\ & = \left[\cos \left(2 \alpha + 2 \beta\right) + i \sin \left(2 \alpha + 2 \beta\right)\right] \times \left(\cos 2 \gamma + i \sin 2 \gamma\right) \\\\ & = \left[\cos \left(2 \alpha + 2 \beta\right) \; \cos 2 \gamma - \sin \left(2 \alpha + 2 \beta\right) \; \sin 2 \gamma\right] \\ & \hspace{1cm} + i \left[\sin \left(2 \alpha + 2 \beta\right) \; \cos 2 \gamma + \cos \left(2 \alpha + 2 \beta \right) \; \sin 2 \gamma\right] \\\\ & = \cos \left(2 \alpha + 2 \beta + 2 \gamma\right) + i \sin \left(2 \alpha + 2 \beta + 2 \gamma\right) \end{aligned}$

$\begin{aligned} \therefore \; \dfrac{a^2 b^2 + c^2}{abc} & = \dfrac{2 \cos \left(2 \alpha + 2 \beta - 2 \gamma\right)\left[\cos \left(2 \alpha + 2 \beta + 2 \gamma\right) + i \sin \left(2 \alpha + 2 \beta + 2 \gamma\right)\right]}{\cos \left(2 \alpha + 2 \beta + 2 \gamma\right) + i \sin \left(2 \alpha + 2 \beta + 2 \gamma\right)} \\\\ & = 2 \cos \left(2 \alpha + 2 \beta - 2 \gamma\right) \\\\ & = 2 \cos 2 \left(\alpha + \beta - \gamma\right) \end{aligned}$

Hence proved.

Complex Numbers

If $x = \cos \alpha + i \sin \alpha$; $\;$ $y = \cos \beta + i \sin \beta$, $\;$ then prove that $\;$ $x^m y^n + \dfrac{1}{x^m y^n} = 2 \cos \left(m \alpha + n \beta\right)$

Given: $\;$ $x = \cos \alpha + i \sin \alpha$; $\;\;$ $y = \cos \beta + i \sin \beta$

$\implies$ $\dfrac{1}{x} = \cos \alpha - i \sin \alpha$; $\;$ $\dfrac{1}{y} = \cos \beta - i \sin \beta$

Now, $\;$ $x^m = \left(\cos \alpha + i \sin \alpha\right)^m = \cos \left(m \alpha\right) + i \sin \left(m \alpha \right)$

$y^n = \left(\cos \beta + i \sin \beta\right)^n = \cos \left(n \beta\right) + i \sin \left(n \beta\right)$

$\begin{aligned} \therefore \; x^m y^n & = \left[\cos \left(m \alpha\right) + i \sin \left(m \alpha\right)\right] \left[\cos \left(n \beta\right) + i \sin \left(n \beta\right)\right] \\\\ & = \left[\cos \left(m \alpha\right) \cos \left(n \beta\right) + i^2 \sin \left(m \alpha\right) \sin \left(n \beta\right)\right] \\ & \hspace{2cm} + i \left[\sin \left(m \alpha\right) \cos \left(n \beta\right) + \sin \left(n \beta\right) \cos \left(m \alpha\right)\right] \\\\ & = \left[\cos \left(m \alpha\right) \cos \left(n \beta\right) - \sin \left(m \alpha\right) \sin \left(n \beta\right)\right] \\ & \hspace{2cm} + i \left[\sin \left(m \alpha\right) \cos \left(n \beta\right) + \sin \left(n \beta\right) \cos \left(m \alpha\right)\right] \\\\ & = \cos \left(m \alpha + n \beta\right) + i \sin \left(m \alpha + n \beta\right) \end{aligned}$

$\dfrac{1}{x^m} = \left(\cos \alpha - i \sin \alpha\right)^m = \cos \left(m \alpha\right) - i \sin \left(m \alpha\right)$

$\dfrac{1}{y^n} = \left(\cos \beta - i \sin \beta\right)^n = \cos \left(n \beta\right) - i \sin \left(n \beta\right)$

$\begin{aligned} \therefore \; \dfrac{1}{x^m y^n} & = \left[\cos \left(m \alpha\right) - i \sin \left(m \alpha\right)\right] \left[\cos \left(n \beta\right) - i \sin \left(n \beta\right)\right] \\\\ & = \left[\cos \left(m \alpha\right) \cos \left(n \beta\right) + i^2 \sin \left(m \alpha\right) \sin \left(n \beta\right)\right] \\ & \hspace{2cm} - i \left[\sin \left(m \alpha\right) \cos \left(n \beta\right) + \sin \left(n \beta\right) \cos \left(m \alpha\right)\right] \\\\ & = \left[\cos \left(m \alpha\right) \cos \left(n \beta\right) - \sin \left(m \alpha\right) \sin \left(n \beta\right)\right] \\ & \hspace{2cm} - i \left[\sin \left(m \alpha\right) \cos \left(n \beta\right) + \sin \left(n \beta\right) \cos \left(m \alpha\right)\right] \\\\ & = \cos \left(m \alpha + n \beta\right) - i \sin \left(m \alpha + n \beta\right) \end{aligned}$

$\begin{aligned} \therefore \; x^m y^n + \dfrac{1}{x^m y^n} & = \cos \left(m \alpha + n \beta\right) + i \sin \left(m \alpha + n \beta\right) \\ & \hspace{2cm} + \cos \left(m \alpha + n \beta\right) - i \sin \left(m \alpha + n \beta\right) \\\\ & = 2 \cos \left(m \alpha + n \beta\right) \end{aligned}$

Hence proved.

Complex Numbers

If $\;$ $x + \dfrac{1}{x} = 2 \cos \theta$, $\;$ prove that $\;$ $x^n + \dfrac{1}{x^n} = 2 \cos \left(n \theta\right)$ $\;$ and $\;$ $x^n - \dfrac{1}{x^n} = 2 i \sin \left(n \theta\right)$

Let $x = \cos \theta + i \sin \theta$

$\begin{aligned} Then, \; \dfrac{1}{x} & = \dfrac{1}{\cos \theta + i \sin \theta} \\\\ & = \dfrac{\cos \theta - i \sin \theta}{\left(\cos \theta + i \sin \theta\right) \left(\cos \theta - i \sin \theta\right)} \\\\ & = \dfrac{\cos \theta - i \sin \theta}{\cos^2 \theta - i^2 \sin^2 \theta} \\\\ & = \cos \theta - i \sin \theta \end{aligned}$

so that $\;$ $x + \dfrac{1}{x} = 2 \cos \theta$

Now,

$x^n = \left(\cos \theta + i \sin \theta\right)^n = \cos \left(n \theta\right) + i \sin \left(n \theta\right)$

$\dfrac{1}{x^n} = \left(\cos \theta - i \sin \theta\right)^n = \cos \left(n \theta\right) - i \sin \left(n \theta\right)$

$\therefore \; x^n + \dfrac{1}{x^n} = \cos \left(n \theta\right) + i \sin \left(n \theta\right) + \cos \left(n \theta\right) - i \sin \left(n \theta\right) = 2 \cos \left(n \theta\right)$

$x^n - \dfrac{1}{x^n} = \cos \left(n \theta\right) + i \sin \left(n \theta\right) - \cos \left(n \theta\right) + i \sin \left(n \theta\right) = 2 i \sin \left(n \theta\right)$

Hence proved.

Complex Numbers

If $\alpha$ and $\beta$ are the roots of the equation $x^2 - 2px + \left(p^2 + q^2\right) = 0$ and $\tan \theta = \dfrac{q}{y + p}$, show that $\dfrac{\left(y + \alpha\right)^n - \left(y + \beta\right)^n}{\alpha - \beta} = q^{n - 1} \left(\dfrac{\sin n\theta}{\sin^n \theta}\right)$

The roots of the given quadratic equation $\;$ $x^2 - 2px + \left(p^2 + q^2\right) = 0$ $\;$ are

$\begin{aligned} x & = \dfrac{2p \pm \sqrt{4p^2 - 4p^2 - 4q^2}}{2} \\\\ & = \dfrac{2p \pm 2iq}{2} \\\\ & = p \pm iq \end{aligned}$

$\because$ $\;$ $\alpha$ and $\beta$ are the roots of the given quadratic equation, let

$\alpha = p + i q$ $\;$ and $\;$ $\beta = p - iq$

Given: $\;$ $\tan \theta = \dfrac{q}{y + p}$

$\implies$ $y = \dfrac{q}{\tan \theta} - p$

$\begin{aligned} \therefore \; \left(y + \alpha\right)^n & = \left(\dfrac{q}{\tan \theta} - p + p + iq\right)^n \\\\ & = \left[q \left(\dfrac{1}{\tan \theta} + i\right)\right]^n \\\\ & = q^n \left[\dfrac{\cos \theta + i \sin \theta}{\sin \theta}\right]^n \\\\ & = \dfrac{q^n \left[\cos \left(n\theta\right) + i \sin \left(n \theta\right)\right]}{\sin^n \theta} \;\;\; \cdots \; (1) \end{aligned}$

$\begin{aligned} \left(y + \beta\right)^n & = \left(\dfrac{q}{\tan \theta} - p + p - iq\right)^n \\\\ & = \left[q \left(\dfrac{1}{\tan \theta} - i\right)\right]^n \\\\ & = q^n \left[\dfrac{\cos \theta - i \sin \theta}{\sin \theta}\right]^n \\\\ & = \dfrac{q^n \left[\cos \left(n\theta\right) - i \sin \left(n \theta\right)\right]}{\sin^n \theta} \;\;\; \cdots \; (2) \end{aligned}$

$\alpha - \beta = p + iq - \left(p - iq\right) = 2 \; i \; q$ $\;\;\; \cdots \; (3)$

$\therefore$ $\;$ We have from equations $(1)$, $(2)$ and $(3)$,

$\begin{aligned} \dfrac{\left(y + \alpha\right)^n - \left(y + \beta\right)^n}{\alpha - \beta} & = \dfrac{\dfrac{q^n \left[\cos \left(n\theta\right) + i \sin \left(n \theta\right)\right]}{\sin^n \theta} - \dfrac{q^n \left[\cos \left(n\theta\right) - i \sin \left(n \theta\right)\right]}{\sin^n \theta}}{2 \;i \;q} \\\\ & = \dfrac{q^n \left[\cos \left(n \theta\right) + i \sin \left(n \theta\right) - \cos \left(n \theta\right) + i \sin \left(n \theta\right)\right]}{2 \;i \;q \; \sin^n \theta} \\\\ & = \dfrac{2 \; i \; q^n \; \sin \left(n \theta\right)}{2 \; i \; q \; \sin^n \theta} \\\\ & = q^{n - 1} \left[\dfrac{\sin \left(n \theta\right)}{\sin^n \theta}\right] \end{aligned}$

Hence proved.

Complex Numbers

Prove that $\left(1 + \cos \theta + i \sin \theta\right)^n + \left(1 + \cos \theta - i \sin \theta\right)^n = 2^{n + 1} \cos^{n} \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right)$, $\;$ $n \in N$

Let $\;$ $z = \cos \theta + i \sin \theta$

$\because$ $\;$ $\left|z\right| = 1$ $\implies$ $\overline{z} = \dfrac{1}{z}$

$\therefore$ $\;$ From equation $(1a)$, $\;$ $\overline{z} = \dfrac{1}{z} = \cos \theta - i \sin \theta$ $\;\;\; \cdots \; (1)$

$\begin{aligned} \therefore \; \left(1 + \cos \theta + i \sin \theta\right)^n + \left(1 + \cos \theta - i \sin \theta\right)^n & = \left(1 + z\right)^n + \left(1 + \dfrac{1}{z}\right)^n \\\\ & = \left(1 + z\right)^n + \dfrac{\left(1 + z\right)^n}{z^n} \\\\ & = \left(1 + z\right)^n \left(1 + \dfrac{1}{z^n}\right) \;\;\; \cdots \; (2) \end{aligned}$

From equation $(1)$,

$\begin{aligned} \dfrac{1}{z^n} & = \left(\cos \theta - i \sin \theta\right)^n \\\\ & = \cos \left(n \theta\right) - i \sin \left(n \theta\right) \;\; [\text{By De Moivre's theorem}] \end{aligned}$

$\begin{aligned} \therefore \; 1 + \dfrac{1}{z^n} & = 1 + \cos \left(n \theta\right) - i \sin \left(n \theta\right) \\\\ & = 2 \cos^2 \left(\dfrac{n \theta}{2}\right) - 2 i \sin \left(\dfrac{n \theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \\\\ & = 2 \cos \left(\dfrac{n \theta}{2}\right) \left[\cos \left(\dfrac{n \theta}{2}\right) - i \sin \left(\dfrac{n \theta}{2}\right)\right] \;\;\; \cdots \; (3a) \end{aligned}$

$\begin{aligned} Now, \; \left(1 + z\right) & = \left(1 + \cos \theta\right) + i \sin \theta \\\\ & = 2 \cos^2 \left(\dfrac{\theta}{2}\right) + 2 i \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right) \\\\ & = 2 \cos \left(\dfrac{\theta}{2}\right) \left[\cos \left(\dfrac{\theta}{2}\right) + i \sin \left(\dfrac{\theta}{2}\right)\right] \end{aligned}$

$\begin{aligned} \therefore \; \left(1 + z\right)^n & = \left\{2 \cos \left(\dfrac{\theta}{2}\right) \left[\cos \left(\dfrac{\theta}{2}\right) + i \sin \left(\dfrac{\theta}{2}\right)\right]\right\}^n \\\\ & = 2^n \cos^n \left(\dfrac{\theta}{2}\right) \left[\cos \left(\dfrac{n \theta}{2}\right) + i \sin \left(\dfrac{n \theta}{2}\right)\right] \;\;\; \cdots \; (3b) \end{aligned}$

$\therefore$ $\;$ We have from equations $(3a)$ and $(3b)$,

$\begin{aligned} \left(1 + z\right)^n \left(1 + \dfrac{1}{z^n}\right) & = 2^n \cos^n \left(\dfrac{\theta}{2}\right) \left[\cos \left(\dfrac{n \theta}{2}\right) + i \sin \left(\dfrac{n \theta}{2}\right)\right] \\ & \hspace{2.5cm} \times 2 \cos \left(\dfrac{n \theta}{2}\right) \left[\cos \left(\dfrac{n \theta}{2}\right) - i \sin \left(\dfrac{n \theta}{2}\right)\right] \\\\ & = 2^{n+1} \cos^n \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \\ & \hspace{1cm} \times \left[\cos \left(\dfrac{n \theta}{2}\right) + i \sin \left(\dfrac{n \theta}{2}\right)\right]\left[\cos \left(\dfrac{n \theta}{2}\right) - i \sin \left(\dfrac{n \theta}{2}\right)\right] \\\\ & = 2^{n+1} \cos^n \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \times \left[\cos^2 \left(\dfrac{n \theta}{2}\right) - i^2 \sin^2 \left(\dfrac{n \theta}{2}\right)\right] \\\\ & = 2^{n+1} \cos^n \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \times \left[\cos^2 \left(\dfrac{n \theta}{2}\right) + \sin^2 \left(\dfrac{n \theta}{2}\right)\right] \\\\ & = 2^{n+1} \cos^n \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \;\;\; \cdots \; (4) \end{aligned}$

$\therefore$ $\;$ From equations $(2)$ and $(4)$ we have,

$\left(1 + \cos \theta + i \sin \theta\right)^n + \left(1 + \cos \theta - i \sin \theta\right)^n = 2^{n + 1} \cos^{n} \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right)$, $\;$ $n \in N$

Hence proved.