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 hasI) No girl.II) At least one boy and one girl.III) At least three girls.

Hint: The given question is related to permutations and combinations. Here we use the formula for the number of ways for selecting $r$ things from $n$ different things.

The following formula will be used to solve the given problem:
The number of ways of selecting $r$ things from $n$ different things is given by $^{n}{{C}_{r}}$ and the value of $^{n}{{C}_{r}}$ is given as $^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$ .
I) For the first case , we have to choose a team of $5$ members from a group of $4$
girls and $7$ boys, such that there are no girls in the team, i.e. all the members should be boys.
So, we have to choose $5$ boys from a group of $7$ boys.
So, number of ways ${{=}^{7}}{{C}_{5}}$
$=\dfrac{7!}{5!(7-5)!}$
$=\dfrac{7!}{5!\times 2!}$
Now, we know, we can write $n!$ as $n(n-1)!=n(n-1)(n-2)!$ . So, $7!=7\times 6\times 5!$
So, the number of ways $=\dfrac{7\times 6\times 5!}{5!\times 2!}=\dfrac{7\times 6}{2}=21$.
Hence , there are $21$ ways of selecting a team of $5$ members from a group of $4$ girls and $7$ boys, such that there are no girls in the team.
II) For the second case, we have to choose a team of $5$ members from a group of $4$
girls and $7$ boys, such that there is at least one girl and one boy in the team.
So, first letâ€™s select one boy and one girl. The number of ways of doing it is given as $^{7}{{C}_{1}}{{\times }^{4}}{{C}_{1}}=7\times 4=28$.
Now, the remaining three members of the team can be girls or boys . So, we have to select three members from a group of nine persons (three girls and six boys).
So, number of ways of selection is given as $^{9}{{C}_{3}}=\dfrac{9!}{3!\times 6!}=\dfrac{9\times 8\times 7}{6}=84$
So, the number of ways of to choosing a team of $5$ members from a group of $4$ girls and $7$ boys, such that there is at least one girl and one boy in the team is given as $^{7}{{C}_{1}}{{\times }^{4}}{{C}_{1}}{{\times }^{9}}{{C}_{3}}=28\times 84=2352$
Hence, there are $2352$ ways of choosing a team of $5$ members from a group of $4$ girls and $7$ boys, such that there is at least one girl and one boy in the team.
III) For the third case, we have to choose a team of $5$ members from a group of $4$
girls and $7$ boys, such that there are at least three girls in the team.
So, first, letâ€™s choose three girls from a group of four girls.
Number of ways of selection is given as $^{4}{{C}_{3}}$.
Now, the remaining two members of the group can be girls or boys. So, we have to select two members from a group of eight persons (one girl and seven boys).
So, the number of ways of selection is given as $^{8}{{C}_{2}}$.
So, the number of ways of choosing a team of $5$ members from a group of $4$ girls and $7$ boys, such that there are at least three girls in the team is given as $^{4}{{C}_{3}}{{\times }^{8}}{{C}_{2}}=\dfrac{4!}{3!\times 1!}\times \dfrac{8!}{6!\times 2!}$
$=4\times \dfrac{8\times 7}{2}=4\times 4\times 7$
$=112$
Hence, there are $112$ ways of choosing a team of $5$ members from a group of $4$ girls and $7$ boys, such that there are at least three girls in the team.

Note: Students generally get confused in the expansions of $^{n}{{C}_{r}}$ and $^{n}{{P}_{r}}$. The expansion of $^{n}{{C}_{r}}$ is given as $^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$ and the expansion of $^{n}{{P}_{r}}$ is given as $^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!}$.