JAMB 2000 · UME · Q49

In how many ways can a delegation of 3 be chosen from among 5 men and 3 women, if at least one man and at least one woman must be included?

Worked solution (try it first)

Two men and one woman: $^5C_2 \times {^3C_1} = 10 \times 3 = 30$ ways.
One man and two women: $^5C_1 \times {^3C_2} = 5 \times 3 = 15$ ways.
These are the only mixes with at least one of each, so add them: $30 + 15 = 45$ , option D.