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How many are the possible outcomes from throwing $n$ (identical) dice.

I know it is a combination with repetition, but don't really know how to apply the formula.

I need an explanatory answer with the process of thinking.

The book's solution:

$$\binom{n + 5}{5}$$

N. F. Taussig
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DoubleOseven
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  • It depends on whether the dice are distinguishable or not. – Peter Franek Mar 25 '16 at 09:07
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    What do you mean exactly? – DoubleOseven Mar 25 '16 at 09:07
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    Try to modify your question in such a way that no one has to guess what you mean. – drhab Mar 25 '16 at 09:08
  • Consider you throw 2 dices. How many outcomes do you have? If they are distinguishable (die1, die2) then clearly 36 options (any number of die 1 and any number of die2). If you cannot distinguish them, then you have less options, as $((1,4)=(4,1))$ etc. – Peter Franek Mar 25 '16 at 09:09
  • n+5 choose 5 is the indistinguishable option. That's just a formula for combinations with repetition. – Peter Franek Mar 25 '16 at 09:12
  • I am trying to visualize it in my head, what indistinguishable means. Could you elaborate? – DoubleOseven Mar 25 '16 at 09:14
  • They are distinguishable if one is red and one is blue, for example. Then you count "$3$ on blue die and $5$ on red die" as a different outcome from "$3$ on the red die and $5$ on the blue die". But if you can't distinguish them, you only see "one $3$ and one $5$" and count it as one option, no matter how it is realized. In your case, you can't distinguish them, as the formula suggests. – Peter Franek Mar 25 '16 at 09:17
  • The question said, identical dice, so I think it's indistinguishable. – user Apr 27 '21 at 12:54

1 Answers1

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A common way to explain it is known as "stars and bars".
I shall illustrate it putting identical balls into distinct bins ($1-6$) depicting the results obtained

One result with $n= 10$, say, would be $\;\;\bullet|\bullet|\bullet|\bullet\bullet\bullet\bullet|\bullet |\bullet\bullet\;\;$

Make two notes: only $5$ dividers are needed to depict $6$ bins, and you could have $0$ balls in some bins, e.g. $||\bullet\bullet\bullet\bullet|\bullet\bullet\bullet|\bullet\bullet\bullet|$ depicting $0-0-4-3-3-0$

So if there are $n$ balls and $k$ bins ($k-1$ dividers), the only choice you have is to place the dividers among the lot, thus

$\dbinom{n+k-1}{k-1}$ which works out to $\dbinom{n+5}{5}$ for your particular example.

You could profitably look here to have a more detailed explanation

true blue anil
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