Question

A committee of 12 members is to be formed from 9 women and 8 men. The number of committees in which the women are in majority is
A) 2720
B) 2702
C) 2270
D) 2278

Answer 1

Hint: In order to solve the above question, first we need to make the pairs of women and men considering all the possibilities such that there should only be 12 members in the committee and the number of women need to be in majority. Later using the commutation formula, we will solve the permutations and we will get the required answer.

Complete step-by-step solution:
We have given that, a committee of 12 members is to be formed from 9 women and 8 men.
It is given in the question that the number of women in a committee needs to be in majority or greater than the number of men in a committee.
Therefore,
Let the number of women be ‘W’ and the number of men be ‘M’.
Thus, the possible permutations are;
(9 women and 3 men), (8 women and 4 men), (7 women and 5 men) i.e.
(9W, 3M), (8W, 4M), (7W, 5M)
Thus,
The total number of ways of forming the committee when the women are in majority is;
\[\Rightarrow \left( {}^{9}{{C}_{9}}\cdot {}^{8}{{C}_{3}} \right)+\left( {}^{9}{{C}_{8}}\cdot {}^{8}{{C}_{4}} \right)+\left( {}^{9}{{C}_{7}}\cdot {}^{8}{{C}_{5}} \right)\]
As we know that,
\[\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Solving the above permutations, we obtained
\[\Rightarrow \left( \dfrac{9!}{9!\left( 9-9 \right)!}\times \dfrac{8!}{3!\left( 8-3 \right)!} \right)+\left( \dfrac{9!}{8!\left( 9-8 \right)!}\times \dfrac{8!}{4!\left( 8-4 \right)!} \right)+\left( \dfrac{9!}{7!\left( 9-7 \right)!}\times \dfrac{8!}{5!\left( 8-5 \right)!} \right)\]
\[\Rightarrow \left( \dfrac{9!}{9!\left( 0 \right)!}\times \dfrac{8!}{3!\left( 5 \right)!} \right)+\left( \dfrac{9!}{8!\left( 1 \right)!}\times \dfrac{8!}{4!\left( 4 \right)!} \right)+\left( \dfrac{9!}{7!\left( 2 \right)!}\times \dfrac{8!}{5!\left( 3 \right)!} \right)\]
On cancelling the common terms from the numerator and denominator we get
\[\Rightarrow \left( \dfrac{8\times 7\times 6}{3\times 2\times 1} \right)+\left( \dfrac{9}{1}\times \dfrac{8\times 7\times 6\times 5}{4\times 3\times 2\times 1} \right)+\left( \dfrac{9\times 8}{2\times 1}\times \dfrac{8\times 7\times 6}{3\times 2\times 1} \right)\]
\[\Rightarrow \left( 56 \right)+\left( 630 \right)+\left( 2016 \right)\]
Adding the numbers, we get
\[\Rightarrow 2702\]
Therefore, 2702 committees can be formed in which the number of women are in the majority.

Hence, the option (B) is the correct answer.

Note: In mathematics, a permutation is known as the choice of ‘r’ things from the given set of ‘n’ things without the replacement. In permutation, order matters.
The formula of permutation is given by;
\[\Rightarrow {}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\]
And a combination is known as the choice of ‘r’ things from the given set of ‘n’ things without the replacement and where the order does not matter.
The formula of combination is given by;
\[\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].