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A committee of 6 is chosen from 8 men and 5 women so as to

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Joined: 23 May 2013
Posts: 189

Kudos [?]: 114 [0], given: 42

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A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]

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New post 12 Nov 2014, 08:37
This is an incredibly easy problem and shows how easy it is to overthink these types of problems.

Let's first think about the total number of committees that can be formed. We have a 6 people, with at least 2 men and 3 women. That leaves one open slot that can be filled by either a man or a woman. We have two possibilities: Either 2 Men and 4 Women, or 3 Men and 3 Women.

Therefore, our total number of committees is: \(C^8_2*C^5_4 + C^8_3 *C^5_3 = \frac{8*7}{2} * 5 + \frac{8*7*6}{3*2} * \frac{5*4}{2} = 28*5 + 56*10 = 140 + 560 = 700\)

We know that the correct answer must be less than 700, because we have to eliminate the committees where the two men server together. The only possibility left is answer A.

Answer: A

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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]

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New post 12 Nov 2014, 13:05
I took the long way, just so I could be conceptually clear (but thanks to the experts for the crystal clear short cuts!)

you can have either 2M, 4 W or 3M and 3W:

Cases where either one of the men is selected:
1C1 * 6C1*5C4 * 2 (two ways in which this can occur)+ 1C1 * 6C2*5C3 *2 (again can occur in two ways)
60+300=360

Cases where neither is chosen: 6C2*5C4 + 6C3 *5C3
75+200=275

TOTAL = 635

A

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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]

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New post 14 Jan 2016, 10:31
Sorry, I am convinced about the first explanation regarding dividing the groups into 3 men*3 women+2 men* 4 women.

However, when confronted with the problem , I calculated it in the other way as stated by many above . The discussion on this other approach has left me confused, Can anyone help me with this to clarify my concepts please ?


So I calculated the number of ways groups can be formed as:
8C2 * 5C3 * 8C1
(men) (women) (remaining bunch as after removing 5 people we have 8 m+w remaining)
**I am assuming it is correct till here. Please correct me if otherwise**

Now I am having trouble in removing the overlap .
I did it as 2C2* 5C3 * 8C2 which is giving a wrong answer.

Thanks

Kudos [?]: 10 [0], given: 5

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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]

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New post 15 Jan 2016, 02:02
shreyashid wrote:
Sorry, I am convinced about the first explanation regarding dividing the groups into 3 men*3 women+2 men* 4 women.

However, when confronted with the problem , I calculated it in the other way as stated by many above . The discussion on this other approach has left me confused, Can anyone help me with this to clarify my concepts please ?


So I calculated the number of ways groups can be formed as:
8C2 * 5C3 * 8C1
(men) (women) (remaining bunch as after removing 5 people we have 8 m+w remaining)
**I am assuming it is correct till here. Please correct me if otherwise**

Now I am having trouble in removing the overlap .
I did it as 2C2* 5C3 * 8C2 which is giving a wrong answer.

Thanks


Hi,
8C2 * 5C3 * 8C1 as a solution for this Q is wrong...

this would be a valid solution if the Q said that you have to choose "A committee of 6 is chosen from 8 men and 5 women so as to contain 2 men and 3 women and one child from a group of 8 childrens"...

the solution 8C2 * 5C3 * 8C1 treats the 8C1 as an independent group..
but you have these 8 people as a part of the other two groups..
so what happens/.. it gives us duplicate scenarios..
I'lljust point to one..
say there are 3 men A,B,C in the group...
A and B are choosen in 8C2 and C is choosen in 8C1..
in another scenario A and C are choosen in 8C2 and B is choosen in 8C1..

so you see both are same case and that is why 8C2 * 5C3 * 8C1 is wrong...

Hope it helps
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]

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New post 10 Sep 2017, 19:38
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For those who, UNDERSTANDABLY, struggled (there is a LOT of calculation in here) to get done with this within 3 mins (aka max time you should spend on any question in GMAT), here is some strategy advice:

Read the question, then glance at answer choices. Now that you have idea of the answer choices, start calculating.

Usually this would go like this:
STEP 1 - Calculate TOTAL committees (WITHOUT restrcitions)
STEP 2 - Calculate the case that IS restricted (put 2 men together)
STEP 3 - Do STEP 1 minus Step 2. You then finally get to the answer... right?

BUTTTTT.....

Hopefully, when you get to the answer of STEP 1, which is 700, we KNOW that when we apply restrictions (700 minus SOME NUMBER), you will end up something less than 700.

ONLY A is less than 700. Boom, done. You dont even need to think about STEP 2. And You certainly dont need to do the math comes with STEP 2.

Keep this in mind for ALL PS questions. GMAT is not about crunching numbers. Leverage all info given in the question, INCLUDING answer choices.

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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]

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New post 19 Oct 2017, 09:35
MY LAZY LOGIC:

The number of possibilities when there are no restrictions of the men who hate each other..

2 Men and 4 Women - 8C2 * 5C4 = 140
3 Men and 3 Women - 8C3 * 5C3 = 560

so total = 700

with the introduction of a restriction, the number of ways will obviously be lesser than 700, only option lesser than 700 is Option A - 635..

hence Option A

Hope you found it helpful, Kudos if so..

Kudos [?]: 11 [0], given: 25

Re: A committee of 6 is chosen from 8 men and 5 women so as to   [#permalink] 19 Oct 2017, 09:35

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