There are 11 women and 9 men in a certain club. If the club is to sele

Bunuel,
Thanks for confirming the ans

11C2 * 9C2 = 11*10/2 * 9*8/2

= 11 * 10 * 9 * 2

= 990 * 2

= 1980

Answer - E

Hi All,

Both barakhaiev and subhashghosh have correctly answer this question, albeit doing the "math" in slightly different ways.

This type of question is relatively rare on the GMAT, but it's more likely to show up when you're scoring at a high level in the Quant section, so the basic ideas behind it are worth memorizing. You have to figure out each individual combination, then multiply the results.

Sometimes the answer choices are "spread out" enough that you can take advantage of that pattern and avoid a little bit of work (and save a little bit of time).

Here, choosing 2 women from a group of 11 is 11c2 = 11!/(2!9!) = (11)(10)/(2)(1) = 55

Choosing 2 men from a group of 9 men is 9c2 = 9!/(2!7!) = (9)(8)/(2)(1) = 36

Now we just have to multiply these two values together: (55)(36).

Since (50)(30) = 1500, and (55)(36) is going to be MUCH LARGER than 1500, there's really only one answer that could possibly be correct.

Final Answer:

Keep the answer choices in mind when you're doing your work; they often provide 'clues' as how best to proceed and potential patterns or shortcuts in how the "math" will work.

GMAT assassins aren't born, they're made,
Rich

We need to determine how many different committees are possible with 2 women and 2 men from 11 women and 9 men. Let’s first determine the number of ways to select 2 women.

number of ways to select 2 women from 11 of them = 11C2 = (11 x 10)/2! = 55 ways

Next we determine the number of ways to select 2 men.

number of ways to select 2 men from 9 of them = 9C2 = (9 x 8)/2! = 36 ways

Thus, the number of ways to select 2 women and 2 men is 55 x 36 = 1,980 ways.

Answer: E

EMPOWERgmatRichC, why don't we divide by 2! when multiplying (55)(36)?
Since the order does not matter, I thought that it would not be needed...

Hi plaverbach,

You're correct that it technically does not matter whether we choose the 'pair' of men first or the 'pair' of women first, BUT we have to consider every combination of 'male pair' and 'female pair' and that's ultimately about multiplication (although you could map out every possible group of 4 - which would be a step-heavy version of addition).

GMAT assassins aren't born, they're made,
Rich

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