permutations

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chaitanyareddy: Senior | Next Rank: 100 Posts; Posts: 30; Joined: Fri Feb 27, 2009 12:45 am; Thanked: 1 times; GMAT Score:620

by chaitanyareddy » Mon Aug 16, 2010 5:46 am

Hi Experts,
Could you please give an explained solution to the following question.

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?

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Source: Beat The GMAT — Problem Solving | Mon Aug 16, 2010 5:46 am

RyanDark: Senior | Next Rank: 100 Posts; Posts: 62; Joined: Fri Oct 23, 2009 12:54 pmFollowed by: 1 member; Location: Pittsburgh,PA; Thanked: 4 times

by RyanDark » Mon Aug 16, 2010 8:11 am

What is the OA? I think the answer I am getting is quite large to be correct.

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clock60: Legendary Member; Posts: 759; Joined: Mon Apr 26, 2010 10:15 amFollowed by: 3 members; Thanked: 85 times

by clock60 » Mon Aug 16, 2010 9:25 am

my bad guys, i actually did not read the problem to the end and so got wrong answer. now it is corrected
2-men,4-women
3-men,3-women

8C2*5C4+8C3*5C3=28*5+56*10=140+560=700- is the number of committee that can be formed without restriction that two men refuse to sit together, now we calculate the number if commttee with exactly two men ( that refuse to sit both) and substract from 700

two men can be picked up the only way-1
4 women from 5 -5 outcomes
so 1 st with 2 men and 4 women 1*5=5 outcomes

now 3 men, two already given, the last - any from 6 left -6 ways
3 women from 5-10 outcomes, and total 1*6*10=60

together 5+60=65, and final result 700-65=635,-target answer

Last edited by clock60 on Mon Aug 16, 2010 12:20 pm, edited 1 time in total.

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chaitanyareddy: Senior | Next Rank: 100 Posts; Posts: 30; Joined: Fri Feb 27, 2009 12:45 am; Thanked: 1 times; GMAT Score:620

by chaitanyareddy » Mon Aug 16, 2010 9:55 am

OA : 635

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RyanDark: Senior | Next Rank: 100 Posts; Posts: 62; Joined: Fri Oct 23, 2009 12:54 pmFollowed by: 1 member; Location: Pittsburgh,PA; Thanked: 4 times

by RyanDark » Mon Aug 16, 2010 10:09 am

I got 360?? OA??

2 types of combinations:
3 Men , 3 Women
-Women can be chosen in 5C3 ways
-Men: since 2 men dont want to be together,lets make them one group.So we need to chose one of those 2 men.So 2C1
And need to choose 2 more men from remaining 6 men.So 6C2
So, 2C1 * 6C2* 5C3=300

2 Men, 4 Women
Women:5C4
Men:Again need to chose only one of the 2 men who dont want to be together,2C1
And choose one out of remaining 6,6C1
So,2C1+6C1*5C4=60

Total =300+60=360

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Gurpinder: Legendary Member; Posts: 659; Joined: Mon Dec 14, 2009 8:12 amFollowed by: 3 members; Thanked: 32 times

by Gurpinder » Mon Aug 16, 2010 10:11 am

chaitanyareddy wrote:Hi Experts,
Could you please give an explained solution to the following question.

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?

Hmm.....this one involves very large numbers. I am not sure if this is a good GMAT problem. Whats the source?

"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.

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crackinggmat: Senior | Next Rank: 100 Posts; Posts: 79; Joined: Thu Jun 03, 2010 11:58 am; Thanked: 1 times

by crackinggmat » Mon Aug 16, 2010 10:21 am

answer is definately 635

to solve we can find total no of ways to find we can do the selection without any restictions as done by clock 60

then we can subtract the number of ways in which the two persons will come together ....which will come 65
hence subtacrt and answer is 635..

i hope this hint would help u guys reach the correct answer

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