Ten student remain in a game of musical chairs. If four chairs ae removed, how many different groups of six students can remain?

um... I think it would be... 6x5x4x3x2x1 divided by 10?? :confused:

To determine if it's a permutation or combination, you need to decide if order matters. If the order matters, then it's permutation. If it does not matter then it's a combination.

For example, if it were a license plate, then ABC is not the same as BAC. Therefore, the order they are in matters. So that's where "order matters" and that's a permutation.

However, if you pick 3 cards and one is an ace, one's a heart and one's a spade, it would not matter what order you picked them in -- you will still have the same 3 cards. So order does not matter and that's a combination.

So, for musical chairs, does order matter? i.e. does it matter which chair they end up in?

Both permutations and combinations are two numbers: the first is the total number of objects you have to pick from. The second number is how many you can choose. i.e. a 3-digit combination lock with single-digit numbers is chosing 3 out of 10 total.

That example is actually a permutation, but let's use the 3 out of 10 to show how they're done:

P(10,3) = \frac{10!}{(10-3)!}

C(10,3) = \frac{10!}{(10-3)!3!}

Notice there's only one difference between the two equations. In both cases the first number is the numerator. In both cases the difference between the 2 numbers (7) is in the denominator. But for a combination we have one more thing: the second number is also in the denominator.

Do you have specific questions about how to work these out?