Question

Team A consists of $7$ boys and $n$ girls and Team B has $4$ boys and $6$ girls. If a total of $52$ single matches can be arranged between these two teams when a boy play against a boy and a girl against a girl, then $n$ is equal to
1. $5$
2. $6$
3. $2$
4. $4$

Answer 1

Hint: By applying the combination formula find the total number of matches between boys of Team A and Team B, then find the matches between girls of Team A and Team B. In the last equate the sum of both matches to $52$.

Formula Used: Combination formula –
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Here, $n = n \times (n - 1) \times (n - 2) \times .......$

Complete step by step Solution:
Given that,
In Team A –
Total number of boys $ = 7$
Total number of girls $ = n$

In Team B –
Total number of boys $ = 4$
Total number of girls $ = 6$

Using the Combination formula,
Total number of matches between boys of Team A and Team B
$ = {}^7{C_1} \times {}^4{C_1}$
$ = \dfrac{{7!}}{{1!\left( {7 - 1} \right)!}} \times \dfrac{{4!}}{{1!\left( {4 - 1} \right)!}}$
$ = \dfrac{{7 \times 6!}}{{6!}} \times \dfrac{{4 \times 3!}}{{3!}}$
$ = 28$
Total number of matches between girls of Team A and Team B
$ = {}^n{C_1} \times {}^6{C_1}$
$ = \dfrac{{n!}}{{1!\left( {n - 1} \right)!}} \times \dfrac{{6!}}{{1!\left( {6 - 1} \right)!}}$
$ = \dfrac{{n \times (n - 1)!}}{{(n - 1)!}} \times \dfrac{{6 \times 5!}}{{5!}}$
$ = 6n$

Given, Total number of single matches can be arranged between teams are $52$
$ \Rightarrow 28 + 6n = 52$
$6n = 52 - 28$
$n = \dfrac{{24}}{6}$
$n = 4$
Hence, Number of girls in Team A are $4$.
Which implies that Option (4) is correct.

Therefore, the correct option is 4.

Note: One should always remember that when order does not matter in a probability question then apply the formula of combination otherwise, go for permutation if the order of the given data matters.