IIT-JEE 1994Question 1
A committee of $12$ is to be formed from $9$ women and $8$ men. In how many ways can this be done if at least five women have to be included in a committee? In how many of these committees
(a) the women are in majority?
(b) the men are in majority?
Permutation and Combination
AnswerTotal $6062$; (a) $2702$; (b) $1008$
Solution
Concept. With $w$ women the committee has $12-w$ men; we need $w\ge5$ and $12-w\le8$. Formulas used. (1) ${}^nC_r={}^nC_{n-r}$(2) Multiplication principle; add the casesStep 1 - Cases $w=5..9$. $w=5$: ${}^9C_5\,{}^8C_7=126\cdot8=1008$ $w=6$: ${}^9C_6\,{}^8C_6=84\cdot28=2352$ $w=7$: ${}^9C_7\,{}^8C_5=36\cdot56=2016$ $w=8$: ${}^9C_8\,{}^8C_4=9\cdot70=630$ $w=9$: ${}^9C_9\,{}^8C_3=1\cdot56=56$ Step 2 - Total. $1008+2352+2016+630+56=6062$. Step 3 - (a) Women majority. $w>12-w\Rightarrow w\ge7$: $2016+630+56=2702$. Step 4 - (b) Men majority. $12-w>w\Rightarrow w\le5$, so $w=5$: $1008$. ($w=6$ is a tie.) Answer: Total $6062$; (a) $2702$; (b) $1008$