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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 \(C_3^{10} - C_3^6 = 120 - 20 = 100\).

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

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?).

\(C^3_6\), \(C^6_3\), 6C3 are the same: choosing 3 out of 6. How can we choose 6 out of 3? Those are just different ways of writing the same.
_________________

Could you tell why my solution of "fixing" one women and then choosing two other members out of the nine remaining didn't work please? As I said, I tried to do so using this formula but my solution did not match yours:

# of ways to select one women out of four = 4 # of ways to select two other members out of nine = 36 \(C_4^1 * C_9^2 = 4 * 36 = 144\)

Could you tell why my solution of "fixing" one women and then choosing two other members out of the nine remaining didn't work please? As I said, I tried to do so using this formula but my solution did not match yours:

# 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

The number you get will have duplications.

The number you get will have duplication.

Consider the group of three women {ABC} Say you select women A with 4C1, next you can get women B and C with 9C2: {A}{BC}. Now, with the same method you can get the following group {B}{AC}, which is basically the same group.
_________________

I have another solution. Clearly something is wrong with it. Bunuel, Could you Please tell me why ? For my understanding only .

So if we have to select 3 people from 4 women 6 men ; at least one of them has to be a woman. We can select in 3 ways : 1 woman 2 men : 4C1 X 6C2 2 women 1 man : 4C2 X 6C1 3 women : 4C3 We ad them all : 19+36+4 = 59

I have another solution. Clearly something is wrong with it. Bunuel, Could you Please tell me why ? For my understanding only .

So if we have to select 3 people from 4 women 6 men ; at least one of them has to be a woman. We can select in 3 ways : 1 woman 2 men : 4C1 X 6C2 2 women 1 man : 4C2 X 6C1 3 women : 4C3 We ad them all : 19+36+4 = 59

You're actually right, but your math is wrong. 4C1 = 4 and 6C2 = 15

So you should be adding up 60+36+4 = 100.

However, you should also note that the published solution takes the least amount of calculation (2 choose functions vs 6), meaning that there is much more room for error!