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 plays against a boy and a girl plays against a girl, then $\mathrm{n}$ is equal to :
Correct Option: , 3
Solution:
Total matches between boys of both team
$={ }^{7} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}=28$
Total matches between girls of both
team $={ }^{n} \mathrm{C}_{1}{ }^{6} \mathrm{C}_{1}=6 \mathrm{n}$
Now, $28+6 \mathrm{n}=52$
$\Rightarrow \mathrm{n}=4$
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