Question

A group of students comprises $5$ boys and $n$ girls. If the number of ways , in which a team of $3$ students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is $1750$, then $n$ is equal to
A. 25
B. 28
C. 27
D. 24

Answer 1

Hint: Atleast $1$ boy and $1$ girl in a team of $3$ students : The number of ways $= ({}^5{C_1} \times {}^n{C_2}) + ({}^5{C_2} \times {}^n{C_1})$ which should be equal to $1750$. Use this to find the value of $n$.

Complete step-by-step answer:
It is given that the group of students comprises $5$ boys and $n$ girls.
So, The number of ways in which a team of $3$ students can be selected from this group such that each team consists of at least one boy and at least one girl, is
$=$(number of ways selecting $1$ boy and $2$ girls) $+$(number of ways selecting $2$ and $1$ girl)
Hence,
$\Rightarrow ({}^5{C_1} \times {}^n{C_2}) + ({}^5{C_2} \times {}^n{C_1}) = 1750$ [given]
$\Rightarrow 5 \times \dfrac{{n(n - 1)}}{{2!}} + 10.n = 1750 \\\ \Rightarrow {n^2} + 3n = 700 \\\ \Rightarrow {n^2} + 3n - 700 = 0 \\\ \Rightarrow n = 25 \\\$

Note: It is advisable in such type of questions first find all the possibilities like in a team of $3$ students comprises of at least $1$ boy and at least $1$ girl , so here the possibilities to form a team are one boy and two girl and second one is two boys and one girl. So, to make the solution simpler, find all the possibilities, this can save time.