Permutations and Combinations: Application Problems (7)

easy 2 min read

Question

How many ways can a committee of 5 be formed from 7 men and 6 women, if the committee must include at least 2 women?

Solution — Step by Step

“At least 2 women” means: 2 women + 3 men, 3 women + 2 men, 4 women + 1 man, or 5 women + 0 men.

  • 2W + 3M: (62)(73)=15×35=525\binom{6}{2}\binom{7}{3} = 15 \times 35 = 525
  • 3W + 2M: (63)(72)=20×21=420\binom{6}{3}\binom{7}{2} = 20 \times 21 = 420
  • 4W + 1M: (64)(71)=15×7=105\binom{6}{4}\binom{7}{1} = 15 \times 7 = 105
  • 5W + 0M: (65)(70)=6×1=6\binom{6}{5}\binom{7}{0} = 6 \times 1 = 6
525+420+105+6=1056525 + 420 + 105 + 6 = 1056

Final answer: 10561056 committees.

Why This Works

“At least” or “at most” problems split into mutually exclusive cases. Add the case counts (since the cases don’t overlap). Each case uses the multiplication principle: choose women × choose men.

Alternative Method (Complement)

Total committees of 5 from 13: (135)=1287\binom{13}{5} = 1287.

Subtract cases with fewer than 2 women:

  • 0 women, 5 men: (60)(75)=21\binom{6}{0}\binom{7}{5} = 21
  • 1 woman, 4 men: (61)(74)=6×35=210\binom{6}{1}\binom{7}{4} = 6 \times 35 = 210

128721210=10561287 - 21 - 210 = 1056. Same answer, often faster when “at least” includes most cases.

For “at least kk of something” with many cases, use the complement: total minus (cases with fewer than kk). For “at least kk” with few excluded cases, this is much faster than enumerating.

Students forget the case “5 women, 0 men” thinking the committee must have at least one man. The problem says “at least 2 women” with no constraint on men, so all-women committees count. Read the constraint carefully.

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