didieravoaka wrote:A student committee that must consist of 5 members is to be formed from a pool of 8 candidates. How many different committees are possible?
(A) 5
(B) 8
(C) 40
(D) 56
(E) 336
You can look at it this way.
You start by selecting one of the 8 for one of the positions in the committee.
You have seven left.
Then you select one of the 7 for the next position.
Then one of the remaining 6 for the third position, one of the remaining 5 for the fourth position, and one of the remaining 4 for the fifth position.
So it seems that you have these numbers of possibilities for each position on the committee.
First: 8
Second: 7
Third: 6
Fourth: 5
Fifth: 4
If the work were done, then we would have 8 x 7 x 6 x 5 x 4 possible different committees.There's an issue though.
The way this math is set up, we can create multiple committees that actually include the same students.
For instance, a certain student. let's call him Danny, could be chosen first, and then a certain other four students chosen. Alternatively, the other four students could be chosen first and then Danny could be chosen last. So there are multiple ways in which the same committee could be chosen, and we have to account for that situation.
The way to account for that is to divide 8 x 7 x 6 x 5 x 4 by the number of ways 5 students can be arranged. By doing that, we account for the fact that each unique committee can be actually chosen as many ways as 5 students can be arranged.
5 students can be arranged 5 x 4 x 3 x 2 x 1 ways.
So the number of possible different committees is (8 x 7 x 6 x 5 x 4)/(5 x 4 x 3 x 2 x 1) = 56
The correct answer is
D.
