Bunuel wrote:

Official Solution:

If 4 women and 6 men work in the accounting department, in how many ways can a committee of 3 be formed if it has to include at least one woman?

A. 36

B. 60

C. 72

D. 80

E. 100

Consider an unconstrained version of the question: in how many ways can a committee of 3 be formed? The answer is \(C_3^{10}\). From this figure we have to subtract the number of committees that consist entirely of men: \(C_3^6\). The final answer is \(C_3^{10} - C_3^6 = 120 - 20 = 100\).

Answer: E

Hello Bunuel,

I think that there is an error with the way you edited the combinaisions which are inverted meaning that you're picking 6 men out 3 (\(C_3^6\)) which is impossible (am I right?).

Also I would have liked to know if there was another way of solving this problem by "fixing" one women and then choosing two other members out of the nine remaining. I tried to do so using this formula but I didn't get the right solution:

# of ways to select one women out of four = 4

# of ways to select two other members out of nine = 36

\(C_4^1 * C_9^2 = 4 * 36 = 144\)

Thank you in advance for your answer,

Carnivor646