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You throw five identical six-sided dice and write down the values showing, in nondecreasing order from left to right. For example, $22245$ means you rolled three $2$s, one $4$, and one $5$. How many outcomes are possible? How many in which all the values are different?

My first instinct is to say that there are $6^5$ such words with repetition and $6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot$ words without repetition. But it turns out that these "words" are actually sets so the solutions are $6 \text { multichoose } 5$ and $6 \text { choose } 5$, respectively.

What terms in the statement of the problem point to the fact that we are counting sets, not words?

permutation
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  • How many in which all the values are different: $\binom65=6$. – barak manos Dec 27 '14 at 16:09
  • “…write down the values showing, in nondecreasing order” means that rolling 42522 yields the same result as rolling 22245. – MJD Dec 27 '14 at 17:41
  • Technically the outcomes are multisets since we count ${2,2,2,4,5}$ and ${2,2,4,4,5}$ as two distinct outcomes. – David K Dec 27 '14 at 17:47

2 Answers2

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The values are put into ascending order from left to right, so 22345 is the same outcome as 22435, since 22435 will be rearranged into 22345. Therefore, the order of the numbers doesn't matter, just like in a set.

Dasherman
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  • The question is not specifically about $22345$. – barak manos Dec 27 '14 at 16:12
  • 22345 and 22435 are just examples, to show that the order doesn't matter. – Dasherman Dec 27 '14 at 16:13
  • @Dasherman, if the values are put into ascending order, how can $22345$ be the same as $22435$? Also, wouldn't ascending order preclude $22435$ from happening since $4 > 3$, for example? – permutation Dec 27 '14 at 16:25
  • Exactly, they are the same outcomes, since 22435 would be precluded (by being turned into 22345 as it were). That is why we are counting sets. The set {2, 2, 3, 4, 5}={2, 2, 4, 3, 5}. This means that the solution isn't $6^5$ because that would mean counting both 22345 and 22435 as separate outcomes. – Dasherman Dec 27 '14 at 16:26
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This is a classic stars and bars problem. If you have ten spaces and fill five of them with bars, then the others - stars - represent the outcomes of the throws. Reading from left to right the value starts at $1$, every time you pass a bar the value goes up by $1$.

Eg $*|*|*|*|*|$ translates to $12345$ while $||***|||**$ represents $33366$. The number of possible outcomes is then $\binom {10}5$. You might want to prove more fully that the translation between interpretations works both ways and doesn't leave anything out.

What matters in your problem, as is clear from the arrangement of outcomes into a standard form from which duplicates can readily be identified, is the number of twos which are thrown, not whether the two twos you get come first and third or fourth and fifth. The "words" retain the order, and count duplicates more than once.

Mark Bennet
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  • But I still don't see what part of the text in the OP points to the fact that order doesn't matter. I am having hell of time seeing why order doesn't matter here :) This is the most difficult part of the problem to me. – permutation Dec 27 '14 at 16:50
  • @permutation It tells you to arrange the scores "in nondecreasing order", which is a canonical order, rather than in the order the numbers originally came in. So $42522$ and $25224$ would both come out written as specified in the question as $22245$. – Mark Bennet Dec 27 '14 at 16:57
  • How would the problem be most likely stated if the order mattered? – permutation Dec 27 '14 at 17:48
  • @permutation There will usually be some indication that the dice are distinguishable, or reference to five throws in order. There aren't universal conventions for these things - you just have to get used to looking out for distinctions. The order is one distinction you need to be alert to every time. The other main one is whether you are allowed duplicates (as here, you can have $2$ three times) - sometimes referred to as "with replacement" - or not, when all the outcomes have to be different ("without replacement"). – Mark Bennet Dec 27 '14 at 21:14