A committee of 4 is to be chosen from 7 employees for a spec : GMAT Problem Solving (PS)
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# A committee of 4 is to be chosen from 7 employees for a spec

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A committee of 4 is to be chosen from 7 employees for a spec [#permalink]

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26 Jan 2012, 16:04
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A committee of 4 is to be chosen from 7 employees for a special project at ACME corportation. 2 of the 7 employees are unwilling to work with each other. How many committes are possible if the 2 employees do not work together?

A. I understand how they got the answer, but not the logic. Please follow my logic and see if I make sense or no sense.

This problem is a combination with restrictions.
My approach: Find the Total Arrangements w/o restrictions and then SUBTRACT the arrangements w/ restrictions.

Group of 4 unordered arrangements out of 7 employees.
N=7 and K = 4; 7!/ (4! * (7-4)!) = 35 groups w/o restrictions.

Next I want to take the 35 groups w/o restriction and SUBTRACT the group w/ Restriction - the group with"Two enemy" employees

There are 7 employees = A B C D E F G So Two of the employees are enemy's thus leaving N = 5 and K is where I'm undecided.
Can you say K = 2 enemy spots.
Basically, N=5 where there are 2 enemys and K=2 where there are 2 enemy spots to fill.

I understand the 5!/(2!*(5-2)! = 10 and then subtract that form 35 and you get 25 as the answer of possible committees w/ restriciton. It's just hard for me to conceptualize the 2nd part w/ restriction correctly on how to think about it.
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Re: A committee of 4 is to be chosen from 7 employees for a spec [#permalink]

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26 Jan 2012, 16:29
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Professor5180 wrote:
A committee of 4 is to be chosen from 7 employees for a special project at ACME corportation. 2 of the 7 employees are unwilling to work with each other. How many committes are possible if the 2 employees do not work together?

Basically there are two approaches possible to solve this problem. Hope that they help to clear your doubts and you'll understand the concept better.

Approach #1:

{# of committees}={total}-{restriction}.

Now, total # of different committees of 4 out 7 people is $$C^4_7=\frac{7!}{4!*3!}=35$$;

# of committees with both A and B in them is $$C^1_1*C^2_5=1*\frac{5!}{3!*2!}=10$$, where $$C^1_1$$ is # of ways to choose A and B out of A and B, which is obviously 1 way to choose, and $$C^2_5=\frac{5!}{3!*2!}$$ is # of ways to choose other 2 people from 7-2=5 people left (I think this was the part you had a problem with);

So, # of committees possible is 35-10=25.

Approach #2:

Direct way: {# of committees}={committees without A and B}+{committees with either A or B}.

# of committees without A and B is $$C^4_5=5$$, where $$C^4_5$$ is # of ways to choose 4 people out of 5 (so without A and B);
# committees with either A or B (but not both) is $$C^1_2*C^3_5=20$$, where $$C^1_2$$ is # of ways to choose either A or B from A and B, and $$C^3_5$$ is # of ways to choose other 3 members of the commitees from 5 people left (7-A-B=5);

So, # of committees possible is 5+20=25.

Similar problem: anthony-and-michael-sit-on-the-six-member-board-of-directors-102027.html

Hope it helps.
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Kudos [?]: 96538 [1] , given: 10753

Re: A committee of 4 is to be chosen from 7 employees for a spec [#permalink]

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27 Jan 2012, 04:43
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mohan514 wrote:
can anyone suggest me some good material for developing my basics,,,

Try Combinatorics chapter of Math Book to have an idea about the staff that is tested on the GMAT: math-combinatorics-87345.html

Also try some questions on combinations to practice:
DS: search.php?search_id=tag&tag_id=31
PS: search.php?search_id=tag&tag_id=52
Hard questions on combinations and probability with detailed solutions: hardest-area-questions-probability-and-combinations-101361.html

Hope it helps.
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Re: A committee of 4 is to be chosen from 7 employees for a spec [#permalink]

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27 Jan 2012, 04:17
can anyone suggest me some good material for developing my basics,,,
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Re: A committee of 4 is to be chosen from 7 employees for a spec [#permalink]

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21 Oct 2014, 14:42
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Re: A committee of 4 is to be chosen from 7 employees for a spec   [#permalink] 21 Oct 2014, 14:42
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