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If four identitical dice are rolled, how many different outcomes can be recorded?

Soln: So I am aware that the solution to this follows a bars and stars sort of procedure and I beleive the solution is C(4+6 -1, 4) = C (4+6 - 1, 5) depending if you are arranging by slots or by bars. Since I am still new to the bars and stars method I am still trying to visualize how the bars and stara would be arranged in a sample outcome. Here is where I am having trouble convincing myself.

So I am dealing with 4 rolls and each dice has 6 possible results so how would a sample outcome look? would it look like this: $$ ****** | ****** | ****** | ******$$

where each bar is a separator for a roll and the stara represemt the numbered result? It doesn't feel like that is how it should be considered

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    $\begingroup$ Do you consider, say $(1,2,2,3)$ as the same outcome as e.g. $(1,3,2,2)$ or as different outcomes? Sorry this is not clear to me. $\endgroup$
    – Jimmy R.
    Commented Dec 12, 2015 at 14:17
  • $\begingroup$ i believe the same $\endgroup$
    – D.C. the III
    Commented Dec 12, 2015 at 14:18
  • $\begingroup$ What is outcome for you? The sum of the 4 numbers, the 4-tuple of the 4 numbers or the different possible combinations? Any of these is related to stars and bars. $\endgroup$
    – user173262
    Commented Dec 12, 2015 at 14:20
  • $\begingroup$ the way I am interpreting it is as the different outcomes possible. But the 4- tuple may be something I will try after $\endgroup$
    – D.C. the III
    Commented Dec 12, 2015 at 14:27
  • $\begingroup$ Oh, yes... my mistake. This is a star and bars problems, yes. You must "invert" the groups to represent it, i.e. put 6 stars to be divided by 4 bars where the bars can stay together in the same hole. $\endgroup$
    – user173262
    Commented Dec 12, 2015 at 14:38

1 Answer 1

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Hint: In this problem, bars should separate the possible values a die roll can have (so $5$ bars), and stars should be the dice ($4$ stars).

So for instance $$||*|**||*$$ corresponds to $3, 4, 4, 6$ on the dice.

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  • $\begingroup$ Tge trickiness really lies in trying to develop those multisets...... sigh. $\endgroup$
    – D.C. the III
    Commented Dec 12, 2015 at 14:52
  • $\begingroup$ @dc3rd: It should get better with practice! $\endgroup$
    – paw88789
    Commented Dec 12, 2015 at 15:00
  • $\begingroup$ Practice.... i enjoy,....... learning it under the time constraint of an exam?.... takes the fun out of it... lol $\endgroup$
    – D.C. the III
    Commented Dec 12, 2015 at 15:12

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