Question

A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has at least one boy and one girl?

Answer 1

Hint: Consider all the possible combinations of choosing 5 people such that it contains at least one boy and one girl. Count the number of ways to choose people for each of the ways using the fact that the number of ways to choose ‘r’ objects from ‘n’ objects is ${}^{n}{{C}_{r}}$. Add up the numbers for all the possible cases.

Complete step-by-step solution -
We have a group of 4 girls and 7 boys. We have to choose a team of 5 members such that it contains at least one boy and one girl.
There are multiple ways to choose 5 members of the team such that it contains at least one boy and one girl. We can choose 4 girls with 1 boy, 3 girls with 2 boys, 2 girls with 3 boys and 1 girl with 4 boys.
We will now calculate the number of ways to choose people in each of the cases.
We know that we can choose ‘r’ objects out of ‘n’ objects in ${}^{n}{{C}_{r}}$ ways.
We will first count the number of ways of choosing 4 girls and 1 boy out of 4 girls and 7 boys. We can choose 4 girls in ${}^{4}{{C}_{4}}$ ways. We can choose 1 boy in ${}^{7}{{C}_{1}}$ ways. So, the total number of ways to choose 4 girls and 1 boy is $={}^{4}{{C}_{4}}\times {}^{7}{{C}_{1}}=\dfrac{4!}{4!0!}\times \dfrac{7!}{1!6!}=1\times 7=7$.

We will now count the number of ways of choosing 3 girls and 2 boys out of 4 girls and 7 boys. We can choose 3 girls in ${}^{4}{{C}_{3}}$ ways. We can choose 2 boys in ${}^{7}{{C}_{2}}$ ways. So, the total number of ways to choose 3 girls and 2 boys is $={}^{4}{{C}_{3}}\times {}^{7}{{C}_{2}}=\dfrac{4!}{3!1!}\times \dfrac{7!}{2!5!}=4\times 21=84$.

We will count the number of ways of choosing 2 girls and 3 boys out of 4 girls and 7 boys. We can choose 2 girls in ${}^{4}{{C}_{2}}$ ways. We can choose 3 boys in ${}^{7}{{C}_{3}}$ ways. So, the total number of ways to choose 2 girls and 3 boys is $={}^{4}{{C}_{2}}\times {}^{7}{{C}_{3}}=\dfrac{4!}{2!2!}\times \dfrac{7!}{4!3!}=6\times 35=210$.

We will count the number of ways of choosing 1 girl and 4 boys out of 4 girls and 7 boys. We can choose 1 girl in ${}^{4}{{C}_{1}}$ ways. We can choose 4 boys in ${}^{7}{{C}_{4}}$ ways. So, the total number of ways to choose 1 girl and 4 boys are $={}^{4}{{C}_{1}}\times {}^{7}{{C}_{4}}=\dfrac{4!}{1!3!}\times \dfrac{7!}{4!3!}=4\times 35=140$.

To calculate the total number of ways to choose a team of 5 members such that it contains at least one boy and one girl, we will add up the number of ways to choose people in each of the cases.
Thus, the number of ways to choose a team of 5 members such that it contains at least one boy and one girl $=7+84+210+140=441$.
Hence, we can choose a team of 5 members such that it contains at least one boy and one girl in 441 ways.

Note: We can also solve this question by calculating the number of ways to choose 5 people out of all the people and the number of ways to choose 5 boys out of 7 boys. Subtract the number of ways to choose 5 boys from the total number of ways to choose 5 people out of all the people to get the number of ways to choose a team of 5 members such that it contains at least one boy and one girl.