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

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.