AP EAMCET · Maths · Permutation Combination

There are two women participating with some men in a chess tournament. Each participant played two games with the other. The number of games that the men played between themselves is 66 more than that of the men played with the women. Then the total number of participants in the tournament is

A \(17\)
B \(13\)
C \(11\)
D \(19\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(13\)

Step-by-step Solution

Detailed explanation

Let there be n men participants Then the number of games that the men play between themselves is \(2{ }^n C_2\) And the number of games that the men played with women is \(2 \times 2 n\) \(\therefore 2{ }^n C_2-2 \times 2 n=66 \Rightarrow \frac{2 n!}{(n-2)!2!}-4 n=66\)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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