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.htmlHope it helps.

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