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

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

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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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Originally posted by JusTLucK04 on 21 Apr 2014, 05:07.
Last edited by Bunuel on 21 Apr 2014, 05:12, edited 1 time in total.
Edited the question and added the OA.
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A committee of four is to be chosen from seven employees for  [#permalink]

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New post 21 Apr 2014, 05:30
4
2
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  [#permalink]

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New post 21 Apr 2014, 05:12
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  [#permalink]

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New post 21 Apr 2014, 05:15
total combinations =7C4
total combinations if 2 persons who dont like each other are selected together= 5C3
Ans= 7C4-5C3
=25
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Re: A committee of four is to be chosen from seven employees for  [#permalink]

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New post 21 Apr 2014, 05:45
Bunuel wrote:
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: anthony-and-michael-sit-on-the-six-member-board-of-directors-102027.html

Hope it helps.


So had 3 employees had a problem with eachother...

7C4 - 3C3*4C1

Highlight a case where the 3C3 portion matters..because it will always be equal to 1 in such cases..
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Re: A committee of four is to be chosen from seven employees for  [#permalink]

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New post 21 Apr 2014, 05:52
JusTLucK04 wrote:
Bunuel wrote:
Bunuel 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: anthony-and-michael-sit-on-the-six-member-board-of-directors-102027.html

Hope it helps.


So had 3 employees had a problem with eachother...

7C4 - 3C3*4C1

Highlight a case where the 3C3 portion matters..because it will always be equal to 1 in such cases..


That's correct.

Yes, nCn=1, and the reason of not omitting this part is just to explain better.
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Re: A committee of four is to be chosen from seven employees for  [#permalink]

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New post 21 Apr 2014, 05:56
Bunuel wrote:
Yes, nCn=1, and the reason of not omitting this part is just to explain better.


So, basically there aren't any variety of questions tested on the GMAT that require playing with this portion of the combination equation

TY
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Re: A committee of four is to be chosen from seven employees for  [#permalink]

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New post 21 Apr 2014, 06:00
JusTLucK04 wrote:
Bunuel wrote:
Yes, nCn=1, and the reason of not omitting this part is just to explain better.


So, basically there aren't any variety of questions tested on the GMAT that require playing with this portion of the combination equation

TY


If you mean by above that we can always omit nCn, then yes,
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Re: A committee of four is to be chosen from seven employees for  [#permalink]

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New post 08 Jun 2016, 10:40
Number of committees with neither of them serving is just the same as forming a committee of 4 from 5 people, which is 5*4*3*2 divided by 4! since the order does not matter, or 5 committees.

To find the number of committees with A but not B, take A and then form the rest of the committee with 3 people from the remaining 5 (not B), so that's 1*(5 choose 3) or 10.

B but not A is the same as above, so total is 5 + 10 + 10 = 25. Answer: C.
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Re: A committee of four is to be chosen from seven employees for  [#permalink]

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New post 25 Feb 2017, 11:06
Bunuel wrote:
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.

Hi Bunuel,

I've made a mistake when calculating the possible number of ways without two persons:
1x1x5x4 = 20 ways without 2 persons, while reviewing I've inferred that I've simply forgotten to unrange possible combinations. Please advise if I correctly identified the root cause of my improper approach. Thanks.
P.S. I'm experiencing difficulties with distinguishing permutation and combinations, and sometimes it ends up with such a foolish mistakes.
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Re: A committee of four is to be chosen from seven employees for  [#permalink]

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