Permutations and Combinations 2 Question 14

14. In a high school, a committee has to be formed from a group of 6 boys $M _1, M _2, M _3, M _4, M _5, M _6$ and 5 girls $G _1$, $G _2, G _3, G _4, G _5$.

(i) Let $\alpha _1$ be the total number of ways in which the committee can be formed such that the committee has 5 members, having exactly 3 boys and 2 girls.

(ii) Let $\alpha _2$ be the total number of ways in which the committee can be formed such that the committee has at least 2 members, and having an equal number of boys and girls.

(iii) Let $\alpha _3$ be the total number of ways in which the committee can be formed such that the committee has 5 members, at least 2 of them being girls.

(iv) Let $\alpha _4$ be the total number of ways in which the committee can be formed such that the committee has 4 members, having at least 2 girls such that both $M _1$ and $G _1$ are NOT in the committee together.

(2018 Adv.)

List-I List-II
P. The value of $\alpha _1$ is 1. 136
Q. The value of $\alpha _2$ is 2. 189
R. The value of $\alpha _3$ is 3. 192
S. The value of $\alpha _4$ is 4. 200
5. 381
6. 461

The correct option is

(a) $P \rightarrow 4 ; Q \rightarrow 6 ; R \rightarrow 2 ; S \rightarrow 1$

(b) $P \rightarrow 1 ; Q \rightarrow 4 ; R \rightarrow 2 ; S \rightarrow 3$

(c) $P \rightarrow 4 ; Q \rightarrow 6 ; R \rightarrow 5 ; S \rightarrow 2$

(d) $P \rightarrow 4 ; Q \rightarrow 2 ; R \rightarrow 3 ; S \rightarrow 1$

Integer Answer Type Question

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Answer:

Correct Answer: 14. (c)

Solution:

  1. Given 6 boys $M _1, M _2, M _3, M _4, M _5, M _6$ and 5 girls $G _1, G _2, G _3, G _4, G _5$

(i) $\alpha _1 \rightarrow$ Total number of ways of selecting 3 boys and 2 girls from 6 boys and 5 girls.

i..e, ${ }^{6} C _3 \times{ }^{5} C _2=20 \times 10=200$

$\therefore \quad \alpha _1=200$

(ii) $\alpha _2 \rightarrow$ Total number of ways selecting at least 2 member and having equal number of boys and girls i.e., ${ }^{6} C _1{ }^{5} C _1+{ }^{6} C _2{ }^{5} C _2+{ }^{6} C _3{ }^{5} C _3+{ }^{6} C _4{ }^{5} C _4+{ }^{6} C _5{ }^{5} C _5$ $=30+150+200+75+6=461$

$\Rightarrow \quad \alpha _2=461$

(iii) $\alpha _3 \rightarrow$ Total number of ways of selecting 5 members in which at least 2 of them girls

$$ \text { i.e., } \begin{aligned} { }^{5} C _2{ }^{6} C _3 & +{ }^{5} C _3{ }^{6} C _2+{ }^{5} C _4{ }^{6} C _1+{ }^{5} C _5{ }^{6} C _0 \\ & =200+150+30+1=381 \\ \alpha _3 & =381 \end{aligned} $$

(iv) $\alpha _4 \rightarrow$ Total number of ways of selecting 4 members in which at least two girls such that $M _1$ and $G _1$ are not included together.

$G _1$ is included $\rightarrow{ }^{4} C _1 \cdot{ }^{5} C _2+{ }^{4} C _2 \cdot{ }^{5} C _1+{ }^{4} C _3$

$$ =40+30+4=74 $$

$M _1$ is included $\rightarrow{ }^{4} C _2 \cdot{ }^{5} C _1+{ }^{4} C _3=30+4=34$

$G _1$ and $M _1$ both are not included

$$ { }^{4} C _4+{ }^{4} C _3 \cdot{ }^{5} C _1+{ }^{4} C _2 \cdot{ }^{5} C _2 $$

$1+20+60=81$

$\therefore$ Total number $=74+34+81=189$

$$ \alpha _4=189 $$

Now, $P \rightarrow 4 ; Q \rightarrow 6 ; R \rightarrow 5 ; S \rightarrow 2$

Hence, option (c) is correct.



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