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
- a particular teacher is included?
- 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
-
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
-
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