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 c

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 and such that both $M_{1}$ and $G_{1}$ are NOT in the committee together.

LIST-I	LIST-II
A. The value of $α_{1}$ is	P. $136$
B. The value of $α_{2}$ is	Q. $189$
C. The value of $α_{3}$ is	R. $192$
D. The value of $α_{4}$ is	S. $200$
	T. $381$
	U. $461$

The correct option is:

TestHub · Accepted Answer

1 α 1 = 6 3 5 2 = 200 So P → 4 2 α 2 = 6 1 5 1 + 6 2 5 2 + 6 3 2 3 + 6 4 5 4 + 6 5 5 5 = 11 5 - 1 = 461 So Q → 6 3 α 3 = 5 2 6 3 + 5 3 6 2 + 5 4 6 1 + 5 5 6 0 = 11 5 - 5 0 6 5 - 5 1 6 4 = 381 So R → 5 4 α 2 = 5 2 6 2 - 4 1 5 1 + 5 3 6 1 - 4 2 1 1 + 5 4 = 189 So S → 2

Mathematics - Permutation & Combination Question with Solution | TestHub

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