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How many 5 person committees chosen at random from a group c

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How many 5 person committees chosen at random from a group c [#permalink] New post 03 Dec 2013, 08:41
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

73% (03:10) correct 27% (01:50) wrong based on 49 sessions
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?

A. 700
B. 1221
C. 1434
D. 2751
E. 3011


Well i know this may look easy. but for me it wasn't.
[Reveal] Spoiler: OA
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Re: How many 5 person committees chosen at random from a group c [#permalink] New post 03 Dec 2013, 13:45
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Expert's post
mumbijoh wrote:
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?
a) 700
b) 1221
c) 1434
d) 2751
e) 3011
Well i know this may look easy. but for me it wasn't.

Dear mumbijoh
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?
Mike :-)
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Re: How many 5 person committees chosen at random from a group c [#permalink] New post 03 Dec 2013, 21:41
Hi Mike,
Thank you for the explanation.I was getting lost with choosing the 5 from men and children I thought that I would have to permutate 5 in each case since you could choose to have all 5 men or all 5 women.
But from your explanation I have gathered that how those two are picked is not in question so long as women are still not part of the group.
Thanks a lot!
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Re: How many 5 person committees chosen at random from a group c [#permalink] New post 05 Dec 2013, 21:47
I added all the possible combinations of a 5 person committee with at least one woman:

5C1*10C4 + 5C2*10C3 + 5C3*10C2 + 5C4*10C1 + 5C5*10C0
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Re: How many 5 person committees chosen at random from a group c [#permalink] New post 06 Dec 2013, 10:02
Expert's post
xhimi wrote:
I added all the possible combinations of a 5 person committee with at least one woman:

5C1*10C4 + 5C2*10C3 + 5C3*10C2 + 5C4*10C1 + 5C5*10C0

Dear xhimi,
Your approach gives the correct answer, but in a way, you fell into the trap of the question. Yes, everything about this expression would yield the correct answer, but without a calculator, this calculation would take quite a long time, longer than the 90 seconds or so you have for a GMAT math problem.

The words "at least" is an important hint that a shortcut is possible. Here's a blog that talks about this in probability:
http://magoosh.com/gmat/2012/gmat-math- ... -question/
It's similar in counting problems --- when the problem say "at least", and especially when it says "at least one" (the most typical case), then it's a trap to get drawn into calculating all the separate cases for one and greater. Instead, it's usually very easy to calculate the total (here, 15C5) and simply subtract the one included in the case for zero. You see, the opposite of "at least one" is "none", and that is almost always the key to solution.

Does all this make sense?
Mike :-)
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Mike McGarry
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Re: How many 5 person committees chosen at random from a group c [#permalink] New post 07 Dec 2013, 00:53
mikemcgarry wrote:
xhimi wrote:
I added all the possible combinations of a 5 person committee with at least one woman:

5C1*10C4 + 5C2*10C3 + 5C3*10C2 + 5C4*10C1 + 5C5*10C0

Dear xhimi,
Your approach gives the correct answer, but in a way, you fell into the trap of the question. Yes, everything about this expression would yield the correct answer, but without a calculator, this calculation would take quite a long time, longer than the 90 seconds or so you have for a GMAT math problem.

The words "at least" is an important hint that a shortcut is possible. Here's a blog that talks about this in probability:

It's similar in counting problems --- when the problem say "at least", and especially when it says "at least one" (the most typical case), then it's a trap to get drawn into calculating all the separate cases for one and greater. Instead, it's usually very easy to calculate the total (here, 15C5) and simply subtract the one included in the case for zero. You see, the opposite of "at least one" is "none", and that is almost always the key to solution.

Does all this make sense?
Mike :-)



makes totally sense. i'm just not very secure in probability yet. my approach indeed took me more than 3 minutes. but your explanation and blog is very helpful, thanks! I'm getting there :-D
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Re: How many 5 person committees chosen at random from a group c [#permalink] New post 12 Dec 2013, 09:35
Number of all possible committees of 15 people minus number of committees without women = number of committees with at least one woman
15C5-10C5= answer
i didn't go through all the calculations though, doesn't seem to be potential gmat question
Re: How many 5 person committees chosen at random from a group c   [#permalink] 12 Dec 2013, 09:35
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