gmat800live wrote:

Hi all! I will be grateful if you could help me. First I saw a questions that says, if I have 4 Women and 3 Men and need to make a committee of 2 people, what's the probability the committee will have 2 women?

Could you confirm the three approaches below? Are they all right?

1. One can do (4/7)*(3/6) = 2/7

Correct!

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2. Also one can do #Have2WomenPermutations/#TotalPermutations and I think that's that's (4*3)/(7*6) = 2/7 - This is correct right?

This is slightly more complicated. You got the right answer, but not

quite for the right reason.

4*3 is what you would do if the order did matter. For instance, suppose your four women are named Anne, Beth, Cassidy, and Dora. Here are your 4x3 options:

Anne and Beth

Anne and Cassidy

Anne and Dora

Beth and Anne

Beth and Cassidy

Beth and Dora

Cassidy and Anne

Cassidy and Beth

Cassidy and Dora

Dora and Anne

Dora and Beth

Dora and Cassidy

See how that's a little weird? We counted the 'Anne and Beth' pair twice, we counted the 'Beth and Dora' pair twice, etc. This is what happens when you act as if order matters, but it actually doesn't.

Technically, the number of pairs would be (4 total women)! / (2 women in the pair)!(2 women not in the pair)!

Which reduces to (4x3x2x1)/(2x1x2x1) = 6 pairs.

Similarly, the total number of pairs is (7 total people)!/(2 people in the pair)!(5 people not in the pair)! or 21 pairs.

6/21 = 2/7, which is the same as the answer you got. The lesson is that when you're doing a problem like this, technically, it doesn't matter how you deal with order, as long as you do it the same way in both the numerator and the denominator. But if the problem just asked you how many committees of two women there could be, the answer would be 6, not 12.

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3. Also one can do (#Total-#DoNotHave2WomenPermutations)/#TotalPermutations

Finally, this is definitely not the right way to do this problem! You would technically get the right answer, but you saw how much more complicated it is. This strategy is only efficient when it's easier to count the 'bad' cases than the 'good' cases. In this problem, it's much easier to count the pairs of women than to count 'all of the other different pairs,' as you saw.

Here's a different question, though, where this strategy

would be better:

"How many different committees could be made that include at least one woman?"

In this case, the 'good' cases are cases with either one or two women, but the 'bad' cases are the ones with two men. It's easier to count the pairs of men, so it would be easier, in this case, to count the male-only pairs and subtract.

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Lastly, I had the thought below after finishing and I'd be grateful if anyone could provide clarity

Ok, so what if I want to know the possible combinations of 2 letters of a set WWWWMMM when order doesn't matter? Logically only 3 options are WM/MM/WW. I get it, but what is the formula that will give me that 3 I am seeking using combinations/permutations? This is driving me crazy! It reminds me to MISISSIPI but I am not able to apply similar strategies here.

Here's how the MISSISSIPPI problem works:

"How many ways can

all of the letters of the word Mississippi be rearranged?"

(11 letters)!/(1 M)!(4 S)!(4 I)!(2 P)! = 11!/(1!x4!x4!x2!) = 11x10x9x8x7x6x5x4x3x2x1 / (1 x 4x3x2x1 x 4x3x2x1 x 2x1)

= 11x10x9x8x7x6x5x

~~4x3x2x1~~ / (

~~1~~ x

~~4x3x2x1~~ x 4x3x2x

~~1~~ x 2x

~~1~~)

= 11x10x9x8x7x6x5 / (4x3x2 x 2)

= 11x10x9x8x7x

~~6~~x5 / (4x

~~3x2~~ x 2)

= 11x10x9x8x7x5 / (4 x 2)

= 11x10x9x

~~8~~x7x5 / (

~~4 x 2~~)

= 11x10x9x7x5 = 34,650

Your question isn't really the same as this. You're not trying to rearrange all seven letters of WWWWMMM. Instead, you're asking a slightly different question: "How many different

two-letter words can be made from the letters WWWWMMM"?

As it turns out, this makes the question a lot more complex! I'm not going to go through the math, because the only situation where the GMAT would ask a question like this, would be if you could count all of the possibilities by hand (which is what you did when you listed WW, WM, MW.) There isn't a single, straightforward equation that'll solve every problem like this. So, I wouldn't worry too much about it.

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I want to add that I think of "combinations" as "Permutations - Redundancies". So basically when I do 5C2 I think... there's 5*4 permutations and I need to eliminate 2! redundancies hence 10. Same with arranging a word such as IISSI we have 5! permutations and we can see the redundancies are going to be 3!2!. The inability to find a similar logic to the above problem is what's worrying me.

This is exactly correct. (Well, you should be thinking 'permutations divided by redundancies', not subtraction, but your example shows that you're understanding it correctly.)

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