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JAMB Mathematics 2010 Paper

Question : 45

Total: 49

Solution:

Concept:

This is a combination problem.

We select women and men independently, then multiply the number of ways.

Explanation:

Choose 2 women from 5 women. The number of ways is

{}^{5}C_{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10.

Choose 3 men from 6 men. The number of ways is

{}^{6}C_{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20.

Multiply the two independent selections:

10 \times 20 = 200.

Therefore, the committee can be chosen in

200

different ways.

Answer:

200

(Option B)

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