TITLE: Arranging 3 items in 7 spots
QUESTION [5 upvotes]: I just wanted to make sure my answer to a problem is correct. 
If we have 3 different items, how many ways can we arrange them amongst 7 spots? What if each item was exactly the same and we didn't care about the order of the 3 items?
If we always pick the same item first and so on, we get 7 * 6 * 5 because for the first item we have 7 possible spots then 6 then 5. Then by symmetry we multiply by 3! to account for the number of possible permutations of the items to get 7!3!/4!? Then we could divide by 3! to drop off all permutations of items to get 7!/4! .
Another way I though about it was we have 7 total items, one group of 3 and one of four. 7!/4! would drop all permutations of the group of 4 which would be a sort of parallel to having 3 items in 7 spaces where we care about the order. From that we could drop all permutations of 3! to get 7!/4!3!.
The numbers seem off. Where did I go wrong? I'm guessing somewhere with the 7!3!/4!

REPLY [2 votes]: If you assume the 3 items are same, then it is not possible to order them.
So the answer is simply in how many ways you can select 3 spots out of the 7.
Which is ${7}\choose3$=$\frac{7!}{3!4!}$.