How many combinations from rolling 5 identical dice?

Question

Where, for example:

(1,3,1,4,6) is considered the same outcome as (1,1,3,6,4)

How many total outcomes are there?

Edit 1:

My hunch is that there are:

6 outcomes from choosing 1 dice to be missing. 6*5 = 30 outcomes from choosing 1 number to be the same and one missing from the 5 remaining. 6*5*4 = 120 from choosing 1 to be triple, 1 from 5 to be missing, and 1 from 4 to be missing. 6*5 = 30 from choosing 1 to be the quadruple and 1 from 5 to be remaining. and 6 from choosing 1 to be all the same number.

Which gives 192 in total.

Edit 2:

Thanks JimmyK for correct answer: 252. Obtained by: C(6+6-1, 6-1) = C(10,5)

Can also get the answer by summing these possibilities, of which I missed out loads above:

C(6,1) + C(6,5) + C(6,1)*C(5,3) + C(6,2)*C(4,1) + C(6,1)*C(5,2) + C(6,1)*C(5,1) = 252

Answer 1

Let $x_i$ be the number of dice with the $i$ face showing. Then, the $x_i$ are non-negative integers with $x_1+x_2+x_3+x_4+x_5+x_6 = 5$. This is now the standard Stars and Bars problem. You can simply use the formula given in that Wikipedia article to get the total number of solutions.

Answer 2

You want to know the number of multisubsets:

$$\left(\!\!\!\binom{n}{k}\!\!\!\right)=\binom{n+k-1}{k}=\frac{n^{(k)}}{k!}$$

Where n is the cardinality of the set where you take elements, and k is the lenght of multisubsets. In your case

$$\frac{6^{(5)}}{5!}=\frac{6\cdot7\cdot8\cdot9\cdot10}{2\cdot3\cdot4\cdot5}=7\cdot4\cdot9=252$$