In a badminton competition involving some men and women of a society every person had to play exactl...........

Quantitative Aptitude (CGL) Permutation & Combination

In a badminton competition involving some men and women of a society, every person had to play exactly one game with every other person. It was found that in 36 games both the players were men and in 78 games both the players were women. Find the number of games in which one player was a man and other was a woman.?

127

117

138

146

None of these

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Explanation:

To solve the problem, let's break it down:

- Men's Games: 36 games between men. If there are \( m \) men, the game formula is \(\frac{m(m-1)}{2} = 36\), leading to \( m(m-1) = 72 \). Solving, \( m = 9 \).

- Women's Games: 78 games between women. If there are \( w \) women, the formula is \(\frac{w(w-1)}{2} = 78\), giving \( w(w-1) = 156 \). Solving, \( w = 13 \).

- Total People: 9 men and 13 women, so 22 people in total.

- Total Games: For 22 people, total games are \(\frac{22 \times 21}{2} = 231\).

- Mixed Games: Subtract men's and women's games from total: \(231 - 36 - 78 = 117\).

- Correct Option:

- Option 2: \(\textbf{117}\)