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

Hint: 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
\[\begin{align}
  & \Rightarrow {{N}_{1}}=\dfrac{5!}{2!\left( 5-2 \right)!}\times \dfrac{n!}{1!\left( n-1 \right)!} \\
 & \Rightarrow {{N}_{1}}=\dfrac{5\times 4\times 3!}{2\times 3!}\times \dfrac{n\times \left( n-1 \right)!}{\left( n-1 \right)!} \\
 & \Rightarrow {{N}_{1}}=10n \\
\end{align}\]
(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
\[\begin{align}
  & \Rightarrow {{N}_{2}}=\dfrac{5!}{1!\left( 5-1 \right)!}\times \dfrac{n!}{2!\left( n-2 \right)!} \\
 & \Rightarrow {{N}_{2}}=\dfrac{5\times 4!}{4!}\times \dfrac{n\times \left( n-1 \right)\times \left( n-2 \right)!}{2\times \left( n-1 \right)!} \\
 & \Rightarrow {{N}_{2}}=\dfrac{5n\left( n-1 \right)}{2} \\
\end{align}\]
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
\[\begin{align}
  & \Rightarrow 10n+\dfrac{5n\left( n-1 \right)}{2}=1750 \\
 & \Rightarrow 5{{n}^{2}}-5n+20n=3500 \\
 & \Rightarrow 5{{n}^{2}}+15n-3500=0 \\
& \Rightarrow {{n}^{2}}+3n-700=0 \\
\end{align}\]
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
\[\begin{align}
  & \Rightarrow {{n}^{2}}+28n-25n-700=0 \\
 & \Rightarrow n\left( n+28 \right)-25\left( n+28 \right)=0 \\
 & \Rightarrow \left( n+28 \right)\left( n-25 \right)=0 \\
\end{align}\]
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
\[\begin{align}
  & \Rightarrow n+28=0 \\
 & \Rightarrow n=-28 \\
\end{align}\]
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
\[\begin{align}
  & \Rightarrow n-25=0 \\
 & \Rightarrow n=25 \\
\end{align}\]
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
\[\begin{align}
  & \Rightarrow n=\dfrac{-15\pm \sqrt{{{\left( 15 \right)}^{2}}-4\left( 5 \right)\left( -3500 \right)}}{2\left( 5 \right)} \\
 & \Rightarrow n=\dfrac{-15\pm \sqrt{225+70000}}{10} \\
 & \Rightarrow n=\dfrac{-15\pm 265}{10} \\
\end{align}\]
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
\[\begin{align}
  & \Rightarrow n=\dfrac{-15+265}{10} \\
 & \Rightarrow n=\dfrac{250}{10}=25 \\
\end{align}\]
Therefore, we can conclude that the number of girls is 25
So, option (d) is the correct answer.