A committee of 4 is to be chosen from 9 men and 4 women. In how many ways can it be chosen if it must have at least one woman? (All 13 persons are different.)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 589
“At least one” = all the committees minus the committees with none.
All committees of 4 from 13 persons: \(\displaystyle ^{13}C_4=\frac{13\times12\times11\times10}{4\times3\times2\times1}=715\)
Committees with no woman (4 men from 9): \(\displaystyle ^9C_4=\frac{9\times8\times7\times6}{24}=126\)
At least one woman \(=715-126=589\)
Why the other options are incorrect
Option (2) 715 — Incorrect. This is all the committees, including the 126 with no woman.
Option (3) 880 — Incorrect. This is \(4\times\,^{12}C_3\) (pick one woman, then any 3 others); a committee with two or more women is counted more than once.
Option (4) 126 — Incorrect. This is the number of committees with no woman.