JEE Advanced 2018 · Paper 2 · Mathematics Q16

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MathematicsList match+3 / -1Question 52 of 54

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 and such that both $M_1$ and $G_1$ are NOT in the committee together.

The correct option is:

LIST-I

P
The value of $\alpha_1$ is
Q
The value of $\alpha_2$ is
R
The value of $\alpha_3$ is
S
The value of $\alpha_4$ is

LIST-II

1
136
2
189
3
192
4
200
5
381
6
461

P → 4; Q → 6; R → 2; S → 1

P → 1; Q → 4; R → 2; S → 3

P → 4; Q → 6; R → 5; S → 2

P → 4; Q → 2; R → 3; S → 1

Answer and worked solution

Official answer: C

$\alpha_1$ : $\binom63\binom52=20\times10=200$ , which is 4.
$\alpha_2$ : $\sum_{k=1}^5\binom6k\binom5k=\binom{11}5-1=461$ (Vandermonde, minus the empty committee), which is 6.
$\alpha_3$ : $\binom{11}5-\binom65-\binom51\binom64=462-6-75=381$ , which is 5.
$\alpha_4$ : 4 members with at least 2 girls:
$\binom52\binom62+\binom53\binom61+\binom54=150+60+5=215.$
Remove those containing both $M_1$ and $G_1$ : the other two members come from 4 girls and 5 boys, with at least one girl, giving $\binom92-\binom52=26$ . So $215-26=189$ , which is 2.

P → 4; Q → 6; R → 5; S → 2.

(C)

Topics:Combinations Properties of binomial coefficients

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