How many 5 person committees chosen at random from a group consisting of 5 men, 5 women, and 5 children contain at least 1 woman?
Well i know this may look easy. but for me it wasn't.
I'm happy to help.
This is NOT an easy question. You may find this blog helpful. http://magoosh.com/gmat/2013/difficult- ... -problems/
Let's first think about this conceptually. The two number of 5-person committees we could choose from 15 people is 15C5. Put that on hold. That's the total number.
Of that total number, all of them would include at least one woman EXCEPT the ones that are all men & children. How many 5-person groups can we form from these 10 people (5 men, 5 children)? That would be 10C5. Those would be the only groups that do NOT have at least one woman.
Thus, subtract this second number from the first. Answer = (15C5) - (10C5)
That's conceptually how we get the answer. Now, calculating that without a calculator is a more of a calculation task than the GMAT would typically make folks do. For this reason, I don't think this is a very good GMAT question. If the calculations are very hairy, the GMAT would typically leave all the answer choices in symbolic form, as I found above.
With a calculator, I will tell you:
15C5 = 3003
10C5 = 252
Answer = 3003 -252 = 2751
Once again, there is absolutely no way the GMAT would expect you to do that calculation on your own, with no calculator.
Does all this make sense?
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