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A committee of four is to be chosen from seven employees for a special

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A committee of four is to be chosen from seven employees for a special  [#permalink]

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New post 26 Jan 2012, 17:04
1
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A
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C
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Question Stats:

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A committee of four is to be chosen from seven employees for a special project at ACME Corporation. Two of the seven employees are unwilling to work with each other. How many committees are possible if the two employees do not work together?

(A) 15
(B) 20
(C) 25
(D) 35
(E) 50
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Re: A committee of four is to be chosen from seven employees for a special  [#permalink]

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New post 21 Apr 2014, 06:30
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Bunuel wrote:
JusTLucK04 wrote:
A committee of four is to be chosen from seven employees for a special project at ACME Corporation. Two of the seven employees are unwilling to work with each other. How many committees are possible if the two employees do not work together?

(A) 15
(B) 20
(C) 25
(D) 35
(E) 50


{The total # of committees possible} - {the number of committees with these two people serving together} = \(C^4_7 - C^2_2*C^2_5=35-10=25\).

Answer: C.


To elaborate more:

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 \(C^2_2*C^2_5=1*\frac{5!}{3!*2!}=10\), where \(C^2_2\) 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: http://gmatclub.com/forum/anthony-and-m ... 02027.html

Hope it helps.
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Re: A committee of four is to be chosen from seven employees for a special  [#permalink]

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New post 27 Jan 2012, 05:17
can anyone suggest me some good material for developing my basics,,,
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Re: A committee of four is to be chosen from seven employees for a special  [#permalink]

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New post 27 Jan 2012, 05:43
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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 four is to be chosen from seven employees for a special  [#permalink]

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New post 21 Apr 2014, 06:12
JusTLucK04 wrote:
A committee of four is to be chosen from seven employees for a special project at ACME Corporation. Two of the seven employees are unwilling to work with each other. How many committees are possible if the two employees do not work together?

(A) 15
(B) 20
(C) 25
(D) 35
(E) 50


{The total # of committees possible} - {the number of committees with these two people serving together} = \(C^4_7 - C^2_2*C^2_5=35-10=25\).

Answer: C.
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Re: A committee of four is to be chosen from seven employees for a special  [#permalink]

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New post 29 Aug 2019, 10:06
Professor5180 wrote:
A committee of four is to be chosen from seven employees for a special project at ACME Corporation. Two of the seven employees are unwilling to work with each other. How many committees are possible if the two employees do not work together?

(A) 15
(B) 20
(C) 25
(D) 35
(E) 50



So I did this way. Let's say you are picking 4 members out of 6 people first. \(C^6_4=15\).
Now remove that one guy from the two guys who don't want to work together and bring the other guy. Again pick 4 from 6. \(C^6_4=15\)
Now think about this. You counted selection of 4 people from 5 people (who were all ready to work with anyone) twice. \(C^5_4=5\).
So 15+15-5=25.
IMO C.
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Re: A committee of four is to be chosen from seven employees for a special   [#permalink] 29 Aug 2019, 10:06
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