SOLUTION: Help please on Permutations. Problem: how many ways can 15 students be separated into two groups of 8 and 7? Attempt: Since they will be separated into 2 groups, i tried us

Question 602332: Help please on Permutations.
Problem:
how many ways can 15 students be separated into two groups of 8 and 7?
Attempt:
Since they will be separated into 2 groups, i tried using 15!/7! since the rest (those seven left) will be on the other group, but it does not show the combinations for the other group. And so i am wrong. :/
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
the order that people are placed into a group is not a factor, so this is a combination question (not permutation)

this is kind of a "trick" question ___ if x + y = z ; then zCx = zCy

ALL the students will be assigned to one group or the other
___ the possible combinations of the two groups are concurrent ___ 15C8 is the same as 15C7
___ separating groups of 8 concurrently separates groups of 7
___ for every group of 8; there will automatically be a complementary group of 7

so the answer is 15C8 (or 15C7)
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