How do I make sense of combinations with repetition? I'm a non-math major but have to study a course in probability and statistics for graduate school. I was able to understand permutations with repetition and then permutations without repetition using examples, and then trying to generalize.
But I can't seem to find any examples of combinations with repetition that explain how the formula works, without going into multisets and things like that.
Help would be much appreciated.
 A: one practical example is the following:
Suppose that you and 7 other friends go to a restaurant.
You each want to order a sandwich, and the restaurant offers 3 varieties of sandwiches (A, B and C). Then combinations with repetitions answer the question "How many different orders can you make"? It's not important who orders which sandwich, only the amount of sandwiches of each type ordered.
You can represent each sandwich to be ordered with an X. So you need to count the ways to distribute 8 X's in three groups:
X X X X X X X X
You can add 2 separator marks to distribute the X's in three subsets, and then claim that the first space is for sandwich of type A, the second for sandwich of type B and the third for sandwiches of type C. Some examples of these are
X X X | X X X X | X (3 A sandwiches, 4 B sandwiches, 1 C sandwich)
X X | X X | X X X X (2 A sandwiches, 2 B sandwiches, 4 C sandwich)
| X X X X X X | X X (0 A sandwiches, 6 B sandwiches, 2 C sandwich)
Note that this represents every way to distribute the sandwiches, and thus counting how to arrange the 2 | symbols and the 8 X gives the solution to the problem.
So, if $CR^3_8$ denotes the combinations with repetition that solve this problem, then
$$CR^3_8 = {3 + 8 - 1\choose8}$$
Because that's the ways you can rearrange the X's and the |'s
A: I frame combinations-with-repetition problems as choosing/picking (order doesn't matter) $r$ items from $n$ categories (each category may be repeatedly picked).
For each selection, illustrate the $n$ categories using $(n-1)$ vertical bars as separators, then represent the $r$ selected items by filling $r$ stars in the according categories.
Notice that every selection contains $[(n-1)+r]$ positions filled by stars & bars, and that selections are distinguished by the $r$ stars filling in different combinations of positions (the vertical bars filling in the remaining positions).
Therefore, the total number of selections is the number of ways to choose $r$ positions from $[(n-1)+r]$ positions, i.e., $$\binom{n-1+r}{r}.$$
A: Combinations with repetition is choosing $k$ objects from a basket with $n$ distinct objects that magically replaces every item you choose so that your choices don't reduce. (Only your $k$th choice isn't replaced as the process gets completed)
In effect, we have added an extra $(k-1)$ items to choose from, hence the formula is
$$\dbinom{n+k-1}{n} = \dbinom{n+k-1}{k-1}.$$
