GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 23 Jan 2020, 15:12

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Current Student
Joined: 23 May 2013
Posts: 181
Location: United States
Concentration: Technology, Healthcare
GMAT 1: 760 Q49 V45
GPA: 3.5
A committee of 6 is chosen from 8 men and 5 women so as to  [#permalink]

### Show Tags

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.

Manager
Joined: 30 Mar 2013
Posts: 101
Re: A committee of 6 is chosen from 8 men and 5 women so as to  [#permalink]

### Show Tags

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
Intern
Joined: 03 Jul 2015
Posts: 27
Re: A committee of 6 is chosen from 8 men and 5 women so as to  [#permalink]

### Show Tags

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
Math Expert
Joined: 02 Aug 2009
Posts: 8335
Re: A committee of 6 is chosen from 8 men and 5 women so as to  [#permalink]

### Show Tags

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
_________________
Intern
Joined: 13 Mar 2016
Posts: 28
A committee of 6 is chosen from 8 men and 5 women so as to  [#permalink]

### Show Tags

10 Sep 2017, 19:38
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.
Intern
Joined: 25 Mar 2017
Posts: 29
Location: India
GPA: 3.5
WE: Social Work (Non-Profit and Government)
Re: A committee of 6 is chosen from 8 men and 5 women so as to  [#permalink]

### Show Tags

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..
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4226
Re: A committee of 6 is chosen from 8 men and 5 women so as to  [#permalink]

### Show Tags

04 Feb 2019, 11:43
Top Contributor
Geronimo wrote:
A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

(1) 635
(2) 700
(3) 1404
(4) 2620
(5) 3510

Since the committee must have 6 people, there are two cases that meet the given restrictions:
1) The committee has 2 men and 4 women
2) The committee has 3 men and 3 women

1) The committee has 2 men and 4 women
Since the order in which we select the men and women does not matter, we can use COMBINATIONS
We can select 2 men from 8 men in 8C2 ways (= 28 ways)
We can select 4 women from 5 women in 5C4 ways (= 5 ways)
So, the total number of ways to select 2 men and 4 women = 28 x 5 = 140

2) The committee has 3 men and 3 women
We can select 3 men from 8 men in 8C3 ways (= 56 ways)
We can select 3 women from 5 women in 5C3 ways (= 10 ways)
So, the total number of ways to select 2 men and 4 women = 56 x 10 = 560

So, the TOTAL number of ways to create the 6-person committee = 140 + 560 = 700

Cheers,
Brent
_________________
Test confidently with gmatprepnow.com
Re: A committee of 6 is chosen from 8 men and 5 women so as to   [#permalink] 04 Feb 2019, 11:43

Go to page   Previous    1   2   [ 27 posts ]

Display posts from previous: Sort by