Question

A debate club consists of 6 girls and 4 boys. A team of 4 members to be selected from this club including the selection of a captain (from amongst these 4 members ) for the team. If the team has to include at most one boy, then the number of ways of selecting the team, is

Answer 1

Hint: Here, we have two ways of forming the team. The team can have either 1 boy and 3 girls or 4 girls and no boy. We will use this concept to find the solution.

Complete step-by-step answer:
Since, it is given that the club consists of 6 girls and 4 boys.
We have to select a team of 4 members.
The first way to select this team of 4 members is:

First we can select a boy from the 4 boys because it is given that the team cannot have more than 1 boy. Then from the 6 girls we can select 3 girls.
After this we will select a leader from this team of 4 chosen persons.
The total no of ways of choosing 1 boy from a team of 4 boys is = $ ^{4}{{C}_{1}} $
Number of ways of choosing 3 girls from 6 girls is = $ ^{6}{{C}_{3}} $
The no of ways of choosing a team leader from a group of 4 persons is = $ ^{4}{{C}_{1}} $

So, the total number of ways of forming a team consisting of 1 boy and 3 girls, one of them being the leader is = $ ^{4}{{C}_{1}}{{\times }^{6}}{{C}_{1}}{{\times }^{4}}{{C}_{1}} $
 $ \begin{align}
  & =\dfrac{4!}{1!\times \left( 4-1 \right)!}\times \dfrac{6!}{3!\times \left( 6-3 \right)!}\times \dfrac{4!}{1!\times \left( 4-1 \right)!} \\
 & =\dfrac{4!}{3!}\times \dfrac{6!}{3!\times 3!}\times \dfrac{4!}{3!} \\
 & =\dfrac{4\times 3!}{3!}\times \dfrac{6\times 5\times 4\times 3!}{3!\times 3!}\times \dfrac{4\times 3!}{3!} \\
 & =4\times \dfrac{6\times 5\times 4}{3\times 2\times 1}\times 4 \\
 & =4\times 20\times 4 \\
 & =320 \\
\end{align} $

Now, the second way of forming the team is:
First we can select 4 girls from 6 girls and then we will choose from these 4 girls.
The number of ways of selecting 4 girls from 6 girls is = $ ^{6}{{C}_{4}} $
Number of ways of selecting a leader from these 4 girls is = $ ^{4}{{C}_{1}} $
So, the total number of ways of forming a team consisting of 4 girls, where one of them is a leader is :
  $ \begin{align}
  & {{=}^{6}}{{C}_{4}}{{\times }^{4}}{{C}_{1}} \\
 & =\dfrac{6!}{4!\times \left( 6-4 \right)!}\times \dfrac{4!}{1!\times \left( 4-2 \right)!} \\
 & =\dfrac{6\times 5\times 4!}{4!\times 2!}\times \dfrac{4!}{1!\times 3!} \\
 & =\dfrac{6\times 5}{2}\times 4 \\
 & =60 \\
\end{align} $
Therefore, total number of ways = 320 +60 = 380 ways.
Hence, the team can be selected in 380 ways.

Note: Students should note here that in the equation it is given that we can choose at most 1 boy. It means that there exists a case where the team can’t have any boy.