Need help with permutations and combinations problems A woman has 6 friends each evening, for 5 days, she invites 3 of them so that the same group is never invited twice. How many ways are there to do this? ( Assume that the order in which groups are invited matters.)
Attempt:
I know if I did, 6 choose 3, I would get all the groups in which it assumes for the subsets of all 6 people {a, b, c, d, e, f} that there are 20 unique subsets of 3. 
However for the second part, where no same group is invited twice, should the solution be :
20 x 19 x 18 x 17 x 16, for the 5 days there are, 1 subset is used, leaving 19 unique subsets as possibilities. I'm just not sure.

The number to make groups of 3 people out of 6 is ...?
Then note that if a group is invited on the first evening, it can't be invited for the second, so there will be one less to choose and so on.
Can you solve it on your own now?
