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JAMB 2010 · UTME · Q49

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

Worked solution (try it first)
  1. Two women from five: 5C2=10^5C_2 = 10 ways.
  2. Three men from six: 6C3=20^6C_3 = 20 ways.
  3. Each choice of women goes with each choice of men, so multiply: 10×20=20010 \times 20 = 200, option B.

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