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An organization constructs a committee of 3 people from A, B

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by Max@Math Revolution » Fri Mar 30, 2018 1:01 am
[GMAT math practice question]

An organization constructs a committee of 3 people from A, B, C, D, E, F and G. A and B are relatives, so they cannot both be committee members at the same time. How many different committees can be formed?

A. 25
B. 30
C. 35
D. 40
E. 45
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by GMATinsight » Sat Mar 31, 2018 11:48 pm
Max@Math Revolution wrote:[GMAT math practice question]

An organization constructs a committee of 3 people from A, B, C, D, E, F and G. A and B are relatives, so they cannot both be committee members at the same time. How many different committees can be formed?

A. 25
B. 30
C. 35
D. 40
E. 45
METHOD 1:

Favourable case = All possible committees of 3 members - Committees of 3 with A and B together in it
7C3 = total ways of selecting 3 members out of 7
5C1 = Since A and B are already selected so we need only one more member out of 5 to make the unwanted teams of 3

i.e. Favourable cases = 7C3 - 5C1 = 35-5 = 30

Answer: option B

METHOD 2:
Case 1: A is selected and B is not = 5C2 = 10
Case 2: B is selected and A is not = 5C2 = 10
Case 3: A and B are both not selected = 5C3 = 10
Total Favourable cases = 10+10+10 = 30

Answer: option B
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by Max@Math Revolution » Sun Apr 01, 2018 5:37 pm
=>
The total number of possible committees of three 3 people chosen from 7 people is 7C3 = 35.
However, we need to exclude the committees containing both A and B.
The number of committees containing A, B and one other person is equal to the number of ways of choosing 1 person from the C, D, E, F, and G, which is 5C1 = 5.
Therefore, the total number of committees 35 – 5 = 30.

Therefore, the answer is B.
Answer: B
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by Scott@TargetTestPrep » Wed Apr 04, 2018 4:51 pm
Max@Math Revolution wrote:[GMAT math practice question]

An organization constructs a committee of 3 people from A, B, C, D, E, F and G. A and B are relatives, so they cannot both be committee members at the same time. How many different committees can be formed?

A. 25
B. 30
C. 35
D. 40
E. 45
We can use the formula:

Total number of committees = committees including A and B + committees not including both A and B

Note that since we are choosing 3 people from 7, the total number of committees is:

7C3 = 7!/(4!*3!) = (7*6*5)/(3*2) = 35.

The number of committees including A and B can be found by observing that since A and B already occupy 2 of the 3 available positions, there are 5 choices left for the third person; therefore there are 5 committees including both A and B.

Thus, the number of committees not including both A and B is 35 - 5 = 30.

Answer: B

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