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
Show Answer
Answer:
Correct Answer: 14. (c)
Solution:
- 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.
