Discussion
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
Answer: D
"At least 3 men" among five people splits into three cases.
\[ 3M, 2W: \ {}^{7}C_{3} \times {}^{6}C_{2} = 35 \times 15 = 525 \]
\[ 4M, 1W: \ {}^{7}C_{4} \times {}^{6}C_{1} = 35 \times 6 = 210 \]
\[ 5M, 0W: \ {}^{7}C_{5} = 21 \]
\[ 525 + 210 + 21 = 756 \]
The cases are mutually exclusive, so the counts add. Here the complement would need just as much work, since "fewer than 3 men" also has three cases.
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