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.htmlHope it helps.
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