Question 13 Marks
If a committee of 12 members is to be formed out of 9 women and 8 men in which at least 5 women are required, then find the member of those committees, if
(i) Women should be in majority,
(ii) Men should be in majority.
(i) Women should be in majority,
(ii) Men should be in majority.
Answer
View full question & answer→In forming a committee of 12 members out of 9 women and 8 men, while there must be at least 5 women, then the following situations arise :
Case (i) If there are 9 women and 3 men, then number
$={ }^9 C_9 \times{ }^8 C_3$
Case (ii) If there are 8 women and 4 men, then number
$={ }^9 C_8 \times{ }^8 C_4={ }^9 C_1 \times{ }^8 C_4$
Case (iii) If there are 7 women and 5 men, then number
$={ }^9 C_7 \times{ }^8 C_5={ }^9 C_2 \times{ }^8 C_3$
Case (iv) If there are 6 women and 6 men, then number
$={ }^9 C_6 \times{ }^8 C_6={ }^9 C_3 \times{ }^8 C_2$
Case (v) If there are 5 women and 7 men, then number
$={ }^9 C_5 \times{ }^9 C_7 \times{ }^9 C_4 \times{ }^9 C_1$
$\therefore$ Required number
$\begin{array}{l}={ }^9 C_9 \times{ }^8 C_3+{ }^9 C_1 \times{ }^8 C_4+{ }^9 C_2 \times{ }^8 C_3+{ }^9 C_3 \times{ }^8 C_2 +{ }^9 C_4 \times{ }^8 C_1 \\=1 \times \frac{8.7 .6}{3.2 .1}+9 \times \frac{8.7 .6 .5}{4.3 .2 .1}+\frac{9.8}{2.1} \times \frac{8.7 .6}{3.2 .1}+\frac{9.8 .7}{3.2 .1} \times \frac{8.7}{2.1}+\frac{9.8 .7 .6}{4.3 .2 .1} \times 8 \\=56+630+2016+2352+1008 \\=6062\end{array}$
(i) If there are more females, then required number
$\begin{array}{l}=\text { case }(\text { i })+\text { case }(\text { ii })+\text { case }\text {(iii) } \\=56+630+2016=2702\end{array}$
(ii) If there are more men than women, then the required number
$=\text { case }(v)=1008$
Case (i) If there are 9 women and 3 men, then number
$={ }^9 C_9 \times{ }^8 C_3$
Case (ii) If there are 8 women and 4 men, then number
$={ }^9 C_8 \times{ }^8 C_4={ }^9 C_1 \times{ }^8 C_4$
Case (iii) If there are 7 women and 5 men, then number
$={ }^9 C_7 \times{ }^8 C_5={ }^9 C_2 \times{ }^8 C_3$
Case (iv) If there are 6 women and 6 men, then number
$={ }^9 C_6 \times{ }^8 C_6={ }^9 C_3 \times{ }^8 C_2$
Case (v) If there are 5 women and 7 men, then number
$={ }^9 C_5 \times{ }^9 C_7 \times{ }^9 C_4 \times{ }^9 C_1$
$\therefore$ Required number
$\begin{array}{l}={ }^9 C_9 \times{ }^8 C_3+{ }^9 C_1 \times{ }^8 C_4+{ }^9 C_2 \times{ }^8 C_3+{ }^9 C_3 \times{ }^8 C_2 +{ }^9 C_4 \times{ }^8 C_1 \\=1 \times \frac{8.7 .6}{3.2 .1}+9 \times \frac{8.7 .6 .5}{4.3 .2 .1}+\frac{9.8}{2.1} \times \frac{8.7 .6}{3.2 .1}+\frac{9.8 .7}{3.2 .1} \times \frac{8.7}{2.1}+\frac{9.8 .7 .6}{4.3 .2 .1} \times 8 \\=56+630+2016+2352+1008 \\=6062\end{array}$
(i) If there are more females, then required number
$\begin{array}{l}=\text { case }(\text { i })+\text { case }(\text { ii })+\text { case }\text {(iii) } \\=56+630+2016=2702\end{array}$
(ii) If there are more men than women, then the required number
$=\text { case }(v)=1008$