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26 Jan 2012, 17:04
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
66% (01:36) correct
34%
(01:52)
wrong
based on 739
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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
(A) 15
(B) 20
(C) 25
(D) 35
(E) 50
21 Apr 2014, 06:30
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Bunuel
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: https://gmatclub.com/forum/anthony-and-m ... 02027.html
Hope it helps.
27 Jan 2012, 05:43
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mohan514
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.
