In a high school, a committee has to be formed from a group of 6 boys and 5 girls . (i) Let 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. [2018]

Question

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 $α_{1}$ 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 $α_{2}$ 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 $α_{3}$ 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 $α_{4}$ 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.

	LIST-I		LIST-II
P.	The value of $α_{1}$ is	1.	136
Q.	The value of $α_{2}$ is	2.	189
R.	The value of $α_{3}$ is	3.	192
S.	The value of $α_{4}$ is	4.	200
		5.	381
		6.	461

The correct option is: [2018]

Smart Achievers · Accepted Answer

(3)

$Given 6 boys M_{1}, M_{2}, M_{3}, M_{4}, M_{5}, M_{6} and 5 girls G_{1}, G_{2}, G_{3}, G_{4}, G_{5}$

$(i) α_{1} \to Total number of ways of selecting 3 boys and 2 girls from 6 boys and 5 girls$

i.e., ${}^{6}C_{3} \times {}^{5}C_{2} = 20 \times 10 = 200$ $∴$ $α_{1} = 200$

$(ii) α_{2} \to Total number of ways of selecting at least 2 members having equal number of boys and girls$

i.e., ${}^{6}C_{1} {}^{5}C_{1} + {}^{6}C_{2} {}^{5}C_{2} + {}^{6}C_{3} {}^{5}C_{3} + {}^{6}C_{4} {}^{5}C_{4} + {}^{6}C_{5} {}^{5}C_{5}$

$= 30 + 150 + 200 + 75 + 6 = 461$ $∴$ $α_{2} = 461$

$(iii) α_{3} \to Total number of ways of selecting 5 members in which at least 2 of them are girls$

i.e., ${}^{5}C_{2} {}^{6}C_{3} + {}^{5}C_{3} {}^{6}C_{2} + {}^{5}C_{4} {}^{6}C_{1} + {}^{5}C_{5} {}^{6}C_{0}$

$= 200 + 150 + 30 + 1 = 381$ $∴$ $α_{3} = 381$

$(iv) α_{4} \to Total number of ways for selecting 4 members in which at least two girls such that M_{1} and G_{1} are not included together.$

$G_{1} is included$ $\to$ ${}^{4}C_{1} \cdot {}^{5}C_{2} + {}^{4}C_{2} \cdot {}^{5}C_{1} + {}^{4}C_{3} = 40 + 30 + 4 = 74$

$M_{1} is included$ $\to$ ${}^{4}C_{2} \cdot {}^{5}C_{1} + {}^{4}C_{3} = 30 + 4 = 34$

$G_{1} and M_{1} both are not included$ $= {}^{4}C_{4} + {}^{4}C_{3} \cdot {}^{5}C_{1} + {}^{4}C_{2} \cdot {}^{5}C_{2} = 1 + 20 + 60 = 81$

$∴$ $Total number = 74 + 34 + 81 = 189$ $∴$ $α_{4} = 189$

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

Solution

1 P → 4; Q → 6; R → 2; S → 1

2 P → 1; Q → 4; R → 2; S → 3

3 P → 4; Q → 6; R → 5; S → 2

4 P → 4; Q → 2; R → 3; S → 1

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