There are 8 members on a board of directors. If they musform a subcommittee of 3 members, how many different subcommittees are possible?
A)6 B)512 C)56 D)336
My answer is: 336
Is this correct?
correct.
I respectfully have to disagree with Bob.
When the word "committee" is used in these type of questions, the implication is that order does not matter, so it is a combination, not a permutation.
The number of subcommittees of 3 members from 8 is C(8,3) = 56
i in turn respectfully disagree with Reiny.
from 8 people there are 3 spots to fill and as correctly stated there is no ordering, this means that for the first spot there are 8 people to choose from, the second there are 7 people left to choose from and the third spot there are 6 people left to choose from, making the number of unique subcommittees available 8*7*6=336
But ahh, Jordan, as soon as you say "for the first spot" you are "ordering".
If John, Mary, and Sue make up the subcommittee, then Sue, Mary and John are the same subcommittee.
In your 336 you would have counted them twice already
So in your 336 you have counted
J,M,S
J,S,M
M,J,S
M,S,J
S,J,M
S,M,J as separate members of the set
Had the question stated to find the number of ways are there to find a president, vice president, and secretary from 8 people, then it would have been
8*7*6 = 336
in saying first spot, i am not implying an order i am saying that there are 3 spots to fill, which there are. they would not be counted twice as you are lowering the number of available people left to fill a spot each time. since there are no repeats once somebody fills a spot they will not fill another spot in the same committee.
Actually Reiny was right, it is 56 I just typed it in and it's correct
Yes, your answer is correct. There are 8 members on the board of directors, and you are forming a subcommittee of 3 members. In order to find the number of different subcommittees, you can use the combination formula.
The combination formula is given by:
C(n, r) = n! / (r!(n - r)!)
where n is the total number of items (in this case, the total number of board members) and r is the number of items chosen (in this case, the number of members in the subcommittee).
Plugging in the values, we have:
C(8, 3) = 8! / (3!(8 - 3)!)
= 8! / (3!5!)
= (8 * 7 * 6 * 5!) / (3 * 2 * 1 * 5!)
= (8 * 7 * 6) / (3 * 2 * 1)
= 336
Therefore, there are 336 different possible subcommittees.