M08-04

Hello Bunuel,

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.

Although eliminating one extreme is the fastest way to solve the problem, in this case by subtracting the number of ways in which only men could form the committee, per the official solution, it may be useful to take the time to understand in practice how you could solve this or a similar probability question by using another application of combinations. Namely, consider that a probability is nothing more than a desired subset within a total, or, in mathematical language,

\(P = \frac{desired}{total}\)

In this case, we can work up to the answer by looking at the committees that could contain 1, 2, or 3 women, respectively. In the first case, 4C1 works nicely--from four total women, we could choose one. But that would mean the other two slots would have to be occupied by men, so out of six men, you would need to choose two: 6C2. Since these are dependent actions--the three slots would have to be filled by 1 woman and 2 men in such a scenario--we would need to multiply them: 4C1 * 6C2 (or 4 * 15, which is 60).

Next, we could repeat the process with 2 women instead: 4C2. We need one man to fill the third slot: 6C1. Together, 4C2 * 6C1 = 6 * 6 = 36.

Finally, we could repeat the process with 3 women filling the three slots, bypassing the men altogether: 4C3 = 4.

Altogether, then, we would have 60 or 36 or 4, or 60 + 36 + 4 different ways of choosing the women to fill the slots of the committee under the given conditions. 60 + 36 + 4 = 100.

As also noted in the official solution, there are 10C3 ways in which to choose 3 people from a pool of 10. This becomes our total from the earlier fraction: 10C3 = 120. Written out in full, then, we get

\(P = \frac{(4C1 * 6C2 + 4C2 * 6C1 + 4C3)}{10C3}\)

The probability boils down to

\(\frac{100}{120}\)

I will reiterate: this would not be the most efficient way to solve the question at hand. However, if you understand how such a solution operates, then you might be able to approach another type of question that is also based on combinations or probability with more confidence, and that is the goal.

Good luck with your studies.

- Andrew

I think this is a high-quality question and I agree with explanation.

Where can I learn about solving these type of problems?
I have no idea here to start

Hello, innav, and welcome to GMAT Club. Note that this question is tagged "Combinatorics and Probability." A good place to start for theory can include any of the following:

• ALL YOU NEED FOR QUANT ! ! ! (see section 7 for many useful links)
• The Probability thread from that link
• The Combinatorics thread from that link
• The Ultimate GMAT Quantitative Megathread (see chapters 21 and 22 specifically)

Many people find probability and combinatorics challenging and spend many hours in hopes of mastering it. Although the exam does test the topic, it does not include any more than one or two such questions (sometimes none, according to a few debriefs I have read). This is not to say that you should brush off a whole topic, but that you should prioritize higher-frequency topics and master fundamentals first.

Good luck with your studies.

- Andrew

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.