Permutations and Combinations 3 Question 2
2. A committee of 11 members is to be formed from 8 males and 5 females. If $m$ is the number of ways the committee is formed with at least 6 males and $n$ is the number of ways the committee is formed with atleast 3 females, then
(2019 Main, 9 April I)
(a) $m=n=68$
(b) $m+n=68$
(c) $m=n=78$
(d) $n=m-8$
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Answer:
Correct Answer: 2. (c)
Solution:
- Since there are 8 males and 5 females. Out of these 13 members committee of 11 members is to be formed.
According to the question, $m=$ number of ways when there is at least 6 males
$$ \begin{aligned} & =\left({ }^{8} C _6 \times{ }^{5} C _5\right)+\left({ }^{8} C _7 \times{ }^{5} C _4\right)+\left({ }^{8} C _8 \times{ }^{5} C _3\right) \\ & =(28 \times 1)+(8 \times 5)+(1 \times 10) \\ & =28+40+10=78 \end{aligned} $$
and $n=$ number of ways when there is at least 3 females
$$ \begin{aligned} & =\left({ }^{5} C _3 \times{ }^{8} C _8\right)+\left({ }^{5} C _4 \times C _7\right)+\left({ }^{5} C _5 \times C _6\right) \\ & =10 \times 1+5 \times 8+1 \times 28=78 \end{aligned} $$
So, $m=n=78$
