Correct Answer: Option D

Explanation

Men = 5

women = 3

Since the chairman must be a man,

Then we can have 1 man, 3 women, or 2 men, 2 women, or 3 men and 1 woman, or 4 men and 0 women.

\(^5 C_1\) x \(^3 C_3\) + \(^5 C_2\) x \(^3 C_2\) + \(^5 C_3\) x \(^3 C_1\) + \(^5 C_4\) x \(^3 C_0\)

= \(\frac{5!}{4!1!}\) x \(\frac{3!}{3!}\) + \(\frac{5!}{2!3!}\) x \(\frac{3!}{2!1!}\) + \(\frac{5!}{3!2!}\) x \(\frac{3!}{1!2!}\) + \(\frac{5!}{4!1!}\) x \(\frac{3!}{0!3!}\)

= 5 + 30 + 30 + 5 = 70 ways.

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