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Ignoring order, how many distinguishable outcomes are there from rolling 6 identical dice? Answer = $462$

I tried a variety of ways such as $\frac{6^6}{6!}$ and can't seem to get the answer. Struggling how to incorporate no order and distinguishable at the same time. Please help.

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  • $\begingroup$ Offhand looks like one might need to consider all the possible partitions of 6. $\endgroup$
    – coffeemath
    Commented Feb 16, 2019 at 12:37

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An outcome here is the same as a six-tuple of non-negative integers that sum to $6$, the $i^{th}$ entry telling you how many times $i$ came up as a value.

Stars and Bars tell us that the number of such is $$\binom {6+6-1}6=462$$

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  • $\begingroup$ Could you explain in more detail please, thank you. $\endgroup$
    – user545637
    Commented Feb 16, 2019 at 12:41
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    $\begingroup$ What part is confusing? The link contains a detailed proof of the relevant formula. $\endgroup$
    – lulu
    Commented Feb 16, 2019 at 12:45
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    $\begingroup$ Is the bijection clear? Since the order of the dice doesn't matter, the only thing that distinguishes two different outcomes is the six-tuple. How many $1's$ did you get? How many $2's$ and so on. Thus you just want to count those six-tuples. $\endgroup$
    – lulu
    Commented Feb 16, 2019 at 12:46
  • $\begingroup$ I'm confused as to why it must sum to 6? $\endgroup$
    – user545637
    Commented Feb 16, 2019 at 12:50
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    $\begingroup$ Say you throw and you get the values $\{1,6,1,1,5,6\}$ We note that you got three $1's$, one $5$ and two $6's$. Thus your outcome would be the six-tuple $(3,0,0,0,1,2)$. Note that these sum to the total number of dice, namely $6$. $\endgroup$
    – lulu
    Commented Feb 16, 2019 at 12:54
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Let $x_1,x_2,...,x_6$ indicate the number of $1,2,3,4,5,6$.

Then the problem can be formulated as: $$x_1+x_2+x_3+x_4+x_5+x_6=6, 0\le x_i\le 6.$$

For example, the following outcomes are equivalent: $$111112\equiv 111121\equiv 111211\equiv 112111\equiv 121111 \Rightarrow \\ (x_1,x_2,x_3,x_4,x_5,x_6)=(5,1,0,0,0,0);\\ 111123\equiv 111213\equiv 112113\equiv 121113\equiv \cdots\equiv 321111 \Rightarrow \\ (x_1,x_2,x_3,x_4,x_5,x_6)=(4,1,1,0,0,0);\\$$

Using Stars and Bars method: $${6+6-1\choose 6-1}=462.$$

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