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.
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can anyone suggest me some good material for developing my basics,,,

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.
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{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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So I did this way. Let's say you are picking 4 members out of 6 people first. \(C^6_4=15\).
Now remove that one guy from the two guys who don't want to work together and bring the other guy. Again pick 4 from 6. \(C^6_4=15\)
Now think about this. You counted selection of 4 people from 5 people (who were all ready to work with anyone) twice. \(C^5_4=5\).
So 15+15-5=25.
IMO C.

Since 2 people cannot work with each other out of 7. It means only 5 people can work with 5 people.

5*5 = 25

Please suggest if my approach is right here?

Hi,

I am afraid your approach is not right.

Method 1

There are 7 people to choose from. Out of 7, 2 can not work together. So we can have total number of ways to form committee of 4 and subtract no of ways in which 2 people work together.

7C4 - 5C2. As two people are already chosen, so we have to choose only 2 people.

=25

Method 2

Here 2 people can not work together, then 1 person can be selected from 2 in 2C1 ways and then 3 people can be selected in 5C3 ways
So we have 2C1*5C3 =20
Also if the 2 people are left out then 4 person can be selected in 5C4 ways =5 ways

Total =20+5=25

Hope it helps

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