Question

There are two women participants in a badminton tournament. The number of games the men played between themselves exceeds by 12 the number of games they played with women. If each player played one game with each other, then the number of men in the tournament was

• A. 4
• B. 6
• C. 7
• D. 8
• E. 10

Hint:

Use $\binom{n}{2}$ for pairwise matches.

Answer 1

The Correct Option is D

Concept: Total matches = combinations.

Step 1: Let number of men = $n$
Men vs men: \[ \binom{n}{2} \] Men vs women: \[ 2n \]

Step 2: Given condition
\[ \binom{n}{2} = 2n + 12 \]

Step 3: Solve
\[ \frac{n(n-1)}{2} = 2n + 12 \] \[ n^2 - n = 4n + 24 \Rightarrow n^2 - 5n - 24 = 0 \] \[ n = 8 \; (\text{valid}) \] Final Conclusion:
Option (D)