Question

A team of three persons with at least one boy and at least one girl is to be formed from 5 boys and $'n'$ girls. If the number of such teams is 1750 then the value of $'n'$ is
(a) 24
(b) 28
(c) 27
(d) 25

Answer 1

We solve this problem by using the combinations of all possible cases such that we can form a team of 3 people such that it has at least 1 boy and at least 1 girl.
We use the combinations that is the number of ways of selecting $'r'$ people from $'n'$ people is given as ${}^{n}{{C}_{r}}$ where,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Then we equal the total number of ways to given number that is 1750 to find the value of $'n'$

Complete step-by-step solution
We are given that a team of 3 persons is formed.
We are given that there are 5 boys and $'n'$ girls.
We are also given that the team consists of at least one boy and at least one girl.
Now, let us take all the possibilities where we can get a team of given condition as
(1) Team of 2 boys and 1 girl
(2) Team of 1 boy and 2 girls.
Let us find the number of ways in each possibility.
(1) Team of 2 boys and 1 girl.
Let us assume that the number of ways of forming the team here as ${{N}_{1}}$
We know that the selection of persons is an example of combinations.
We know that the number of ways of selecting $'r'$ people from $'n'$ people is given as ${}^{n}{{C}_{r}}$ where,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
By using the above formula we get the number of ways of selecting 2 boys from 5 boys as ${}^{5}{{C}_{2}}$
Similarly, we get the number of ways of selecting 1 girl from $'n'$ girls is ${}^{n}{{C}_{1}}$
We know that the total number of ways of selecting 2 boys and 1 girl is the permutation of individual event
So, we can say that the total number of ways of forming team of 2 boys and 1 girl is the permutations for selecting 2 boys and 1 girl separately then we get
$\Rightarrow {{N}_{1}}={}^{5}{{C}_{2}}\times {}^{n}{{C}_{1}}$
Now, by using the formula of combinations that is ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ we get

& \Rightarrow {{N}_{1}}=\dfrac{5!}{2!\left( 5-2 \right)!}\times \dfrac{n!}{1!\left( n-1 \right)!} \\\ & \Rightarrow {{N}_{1}}=\dfrac{5\times 4\times 3!}{2\times 3!}\times \dfrac{n\times \left( n-1 \right)!}{\left( n-1 \right)!} \\\ & \Rightarrow {{N}_{1}}=10n \\\ \end{aligned}$$ (2) Team of 1 boy and 2 girls. Let us assume that the number of ways of forming the team here as $${{N}_{2}}$$ We know that the selection of persons is an example of combinations. We know that the number of ways of selecting $$'r'$$ people from $$'n'$$ people is given as $${}^{n}{{C}_{r}}$$ where, $${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$$ By using the above formula we get the number of ways of selecting 1 boy from 5 boys as $${}^{5}{{C}_{1}}$$ Similarly, we get the number of ways of selecting 2 girls from $$'n'$$ girls is $${}^{n}{{C}_{2}}$$ We know that the total number of ways of selecting 1 boy and 2 girls is the permutation of individual event So, we can say that the total number of ways of forming team of 1 boy and 2 girls is the permutations for selecting 1 boy and 2 girls separately then we get $$\Rightarrow {{N}_{2}}={}^{5}{{C}_{1}}\times {}^{n}{{C}_{2}}$$ Now, by using the formula of combinations that is $${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$$ we get $$\begin{aligned} & \Rightarrow {{N}_{2}}=\dfrac{5!}{1!\left( 5-1 \right)!}\times \dfrac{n!}{2!\left( n-2 \right)!} \\\ & \Rightarrow {{N}_{2}}=\dfrac{5\times 4!}{4!}\times \dfrac{n\times \left( n-1 \right)\times \left( n-2 \right)!}{2\times \left( n-1 \right)!} \\\ & \Rightarrow {{N}_{2}}=\dfrac{5n\left( n-1 \right)}{2} \\\ \end{aligned}$$ We are given that the total number of ways as 1750 By converting the above statement in to mathematical equation we get $$\Rightarrow {{N}_{1}}+{{N}_{2}}=1750$$ By substituting the required values in above equation we get $$\begin{aligned} & \Rightarrow 10n+\dfrac{5n\left( n-1 \right)}{2}=1750 \\\ & \Rightarrow 5{{n}^{2}}-5n+20n=3500 \\\ & \Rightarrow 5{{n}^{2}}+15n-3500=0 \\\ & \Rightarrow {{n}^{2}}+3n-700=0 \\\ \end{aligned}$$ Now, let us use the factorisation method that is let us rewrite the middle term in such a way that we can get factors as follows $$\begin{aligned} & \Rightarrow {{n}^{2}}+28n-25n-700=0 \\\ & \Rightarrow n\left( n+28 \right)-25\left( n+28 \right)=0 \\\ & \Rightarrow \left( n+28 \right)\left( n-25 \right)=0 \\\ \end{aligned}$$ We know that is $$a\times b=0$$ then either of $$a$$ or $$b$$ equals to 0 By using the above condition let us take the first term then we get $$\begin{aligned} & \Rightarrow n+28=0 \\\ & \Rightarrow n=-28 \\\ \end{aligned}$$ We know that the number of girls can never be negative So, we can remove this term Now, let us take the second term then we get $$\begin{aligned} & \Rightarrow n-25=0 \\\ & \Rightarrow n=25 \\\ \end{aligned}$$ Therefore, we can conclude that the number of girls is 25 **So, option (d) is the correct answer.** **Note:** We can solve the value of $$n$$ in other methods also. We know that the roots of quadratic equation of form $$a{{x}^{2}}+bx+c=0$$ are given as $$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$ We have the equation of $$n$$ as $$\Rightarrow 5{{n}^{2}}+15n-3500=0$$ By using the above formula of roots of quadratic equation we get $$\begin{aligned} & \Rightarrow n=\dfrac{-15\pm \sqrt{{{\left( 15 \right)}^{2}}-4\left( 5 \right)\left( -3500 \right)}}{2\left( 5 \right)} \\\ & \Rightarrow n=\dfrac{-15\pm \sqrt{225+70000}}{10} \\\ & \Rightarrow n=\dfrac{-15\pm 265}{10} \\\ \end{aligned}$$ Here, we can get two values for $$n$$ taking positive and negative sign We know that the number of girls can never be negative So, by taking the positive sign we get $$\begin{aligned} & \Rightarrow n=\dfrac{-15+265}{10} \\\ & \Rightarrow n=\dfrac{250}{10}=25 \\\ \end{aligned}$$ Therefore, we can conclude that the number of girls is 25 So, option (d) is the correct answer.