Question

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least three men are in committee. In how many ways can it be done?
A. 624
B. 702
C. 756
D. 812

Answer 1

Hint: We will analyze the question and see that for selecting five persons out of which at least 3 should be men three cases are formed, we will write down the three cases and find out the number of ways by applying the following formula: ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. After that, we will add all the values obtained from the three cases to get the answer.

Complete step-by-step solution
We are given a total of 7 men and a total of 6 women. Now, we have to select five persons to form a committee out of which at least 3 should be men, which means the minimum number of men is 3, therefore we get the following three cases in which the committee can be formed.
Let’s take a look at the three cases:
Case I: Here, in the committee of 5 persons we have 3 men and 2 women, therefore the total number of ways that can be done is by selecting 3 men from 7 men and 2 women from 6 women. Now, for this we will apply the combination formula for selecting r objects from a total of n objects: ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
So, the ways in which we select 3 men from 7 men and 2 women from 6 women are as follows:
$\Rightarrow {}^{7}{{C}_{3}}\times {}^{6}{{C}_{2}}=\left( \dfrac{7!}{3!\left( 7-3 \right)!} \right)\times \left( \dfrac{6!}{2!\left( 6-2 \right)!} \right)=35\times 15=525\text{ }......\left( 1 \right)$
Case II: Here, in the committee of 5 persons we have 4 men and 1 woman, therefore the total number of ways that can be done is by selecting 4 men from 7 men and 1 woman from 6 women. Now, again we will apply the combination formula: ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
So, the ways in which we select 3 men from 7 men and 1 woman from 6 women are as follows:
$\Rightarrow {}^{7}{{C}_{4}}\times {}^{6}{{C}_{1}}=\left( \dfrac{7!}{4!\left( 7-4 \right)!} \right)\times \left( \dfrac{6!}{1!\left( 6-1 \right)!} \right)=35\times 6=210\text{ }......\left( 2 \right)$
Case III: Here, in the committee of 5 persons we have 5 men and 0 women, therefore the total number of ways that can be done is by selecting 5 men from 7 men and 0 women from 6 women. Now, again we will apply the combination formula: ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
So, the ways in which we select 5 men from 7 men is as follows:
\[\Rightarrow {}^{7}{{C}_{5}}=\left( \dfrac{7!}{5!\left( 7-5 \right)!} \right)=21\text{ }......\left( 3 \right)\]
Now, we will be adding all these cases to get the total number of ways in which five persons are selected to form a committee so that at least three men are in the committee, therefore from equation 1, 2, and 3: $525+210+21=756$
Hence, the correct answer is C.

Note: Students might get confused between permutations and combinations. Remember that permutation is used while arranging the objections whereas the combination is used for the selection of the objects. Now, if we apply the permutation then we will use the following formula: ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$, but by this formula, we will not get the right answer, for example, if we select 5 men out of 7 men then ${}^{7}{{P}_{5}}=\dfrac{7!}{\left( 7-5 \right)!}=2520$, which is not correct.