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
- exactly $3$ women?
- at least $3$ women?
- at most $3$ women?
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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
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At least $3$ women
i.e. $\;$ the committee can have $3$ OR $4$ women members
- Number of ways of forming a committee with $3$ women members $= 280$ ways
- $4$ 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
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At most $3$ women
i.e. $\;$ the committee can have $3$ OR $2$ OR $1$ OR $0$ women members
- Number of ways of forming a committee with $3$ women members $= 280$ ways
- $2$ 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 - $1$ 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 - When $0$ women members are present i.e. all members are men, then the committee can be formed in ${^{8}}{C}_{7} = 8$ ways