Combinations with duplicates (e.g. the number of ways of distributing 6 unique t-shirts among 4 children)

[timekeeper]

So for the example of the number of ways of distributing 6 unique t-shirts among 4 children, so that each of the children has exactly one t-shirt, I know that one can calculate the number of combinations using the combination formula:

C(6,4) = 6! / ((6-4)! * 4!) = 15

What I don't understand is how this equation would be modified, if in the 6 t-shirts we had 4 unique t-shirts and 2 duplicate t-shirts. How could one calculate the new number of combinations, if one were to distribute those 6 t-shirts (4 unique, 2 duplicate) among 4 children?

 

[pka]

So for the example of the number of ways of distributing 6 unique t-shirts among 4 children, so that each of the children has exactly one t-shirt, I know that one can calculate the number of combinations using the combination formula:
C(6,4) = 6! / ((6-4)! * 4!) = 15

Where did you get the idea that this has any thing to do with combinations?
It does not. It is a permutation question: [imath] \mathscr{P}(6,4)=?[/imath]

 

[Dr.Peterson]

So for the example of the number of ways of distributing 6 unique t-shirts among 4 children, so that each of the children has exactly one t-shirt, I know that one can calculate the number of combinations using the combination formula:

C(6,4) = 6! / ((6-4)! * 4!) = 15

Let's stick with this first example. Please explain your understanding of the meaning of "unique" and "distribute".

If I take it to mean we want to give each of 6 distinguishable shirts (e.g. 6 different colors) to one of 4 distinguishable children (e.g.with names A, B, C, D), I don't see how each can get only one shirt. Are you taking "distribute" to mean that not all have to be given?

Once we clarify the meaning of your words, I hope we can understand the second example. You'll still have to explain what "duplicate" means to you. (If these come from problems that were given to you, please quote them exactly (even if it's in a different language).

 

[blamocur]

So for the example of the number of ways of distributing 6 unique t-shirts among 4 children, so that each of the children has exactly one t-shirt, I know that one can calculate the number of combinations using the combination formula:

C(6,4) = 6! / ((6-4)! * 4!) = 15

What I don't understand is how this equation would be modified, if in the 6 t-shirts we had 4 unique t-shirts and 2 duplicate t-shirts. How could one calculate the new number of combinations, if one were to distribute those 6 t-shirts (4 unique, 2 duplicate) among 4 children?

It looks to me that you've computed the number of ways of picking 4 t-shirts out of the set of 6. But there is still more than one way to distribute those t-shirts to 4 children.