What is the number of outcomes of this die counting problem?

Question

I have this counting problem I would like someone to help me resolve. The question is:

When playing Yahtzee, you roll five regular 6-sided dice. How many different outcomes are possible from a single roll? The order of the dice does not matter.

The way I calculated it, I reckoned that the first dice would give either of 6 outcomes, so for the second to the fifth die. So, I said the outcomes would be 6^5. But the textbook gave the answer as a combination of C(10,5). Can someone explain to me how he went about getting C(10,5)? it baffles me.

Answer 1

To see how to apply stars and bars, let $x_i$ be the number of $i's $ you get when you roll the $5$ dice. So we want the number of 6-tuples that satisfy $x_1+x_2+x_3+x_4+x_5+x_6 = 5$ and each $x_i$ can range from $0$ to $5$. This is standard stars and bars.

Answer 2

The order of the dice does not matter.

What order the 5 dice show the numbers in is of no importance. The only thing that distinguishes one outcome from another is a different number of $1's,2's,3's,4's,5's$ or $6's$. As an example, $(1,2,2,2,4)$ and $(2,2,4,2,1)$ are identical outcomes, as far as the game is concerned.

Let's say the number of dice that showed $i$ in a roll were $x_i$, $1≤i≤6$.

The number of outcomes would be the number of ordered 6-touples $(x_1,x_2,x_3,x_4,x_5,x_6)$ that satisfy the following conditions:

1. Each $x_i$is a non-negative integer

2. $x_1+x_2+x_3+x_4+x_5+x_6=5$, since the number of dice rolled was 5.

Number of solutions of such a system is 10C5, easy enough to calculate. Check https://youtu.be/DQHg_UcS3hQ for help with this calculation.

Answer 3

Following ${\tt @lulu}$ above comment ( "$\ldots$we are counting the number of $6$−tuples of non-negative integers that sum to $5$". ): \begin{align} \sum_{d_{1} = 0}^{6}\ldots\sum_{d_{6} = 0}^{6} \left[z^{5}\right]z^{d_{1}\ +\ \cdots\ +\ d_{6}} & = \left[z^{5}\right]\sum_{d_{1} = 0}^{\infty}\ldots \sum_{d_{6} = 0}^{\infty}z^{d_{1}\ +\ \cdots\ +\ d_{6}} = \left[z^{5}\right]\left(\sum_{d = 0}^{\infty}z^{d}\right)^{6} \\[5mm] & = \left[z^{5}\right]\left(1 - z\right)^{-6} = {-6 \choose 5}\left(-1\right)^{5} = {6 + 5 - 1 \choose 5} \\[5mm] & = {10 \choose 5} =\bbox[10px,#ffd,border:1px groove navy]{\large252} \end{align}