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# 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.

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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.

Complete step-by-step answer:
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 let’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, let’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 $^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!}$.