JAMB 2010 Mathematics · Question 14

In how many ways can a committee of 2 women and 3 men be chosen from 6 men and 5 women?

Options

A.100
B.200
C.30
D.50

Answer: B. 200

Explanation written with AI and checked against the answer key

Analyze the question: We need to choose 2 women from 5 women AND 3 men from 6 men. Order doesn't matter in a committee, so we use combinations. Concept: The number of ways to choose r items from n is nCr = n! / [r!(n − r)!] When two independent choices are made together ("and"), we multiply the results. Solution: Women: 5C2 = (5 × 4) / (2 × 1) = 10 Men: 6C3 = (6 × 5 × 4) / (3 × 2 × 1) = 120 / 6 = 20 Total number of ways = 5C2 × 6C3 = 10 × 20 = 200 Answer: The correct option is B. 200. Tip: In "choose from groups" questions, work out each group separately with nCr and multiply. Use a shortcut like 6C3 = 6 × 5 × 4 / 3! to avoid computing big factorials. Remember: "and" means multiply, "or" means add.

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