It's a classic question: Let's say you have four dice, how many possible outcomes can you have (the order doesn't matter)? However, instead of using the stars and bars method which gives ${9 \choose 4} = 126$ as an answer, I saw this statement which I think should be equivalent to stars and bars method but I can't prove it:
WLOG assume the four dies' outcomes are $(a, b, c, d)$ and they are ordered so $$ 1 \leq a \leq b \leq c \leq d \leq 6. $$ Now let $(a^{\ast}, b^{\ast}, c^{\ast}, d^{\ast})$ be $(a+0, b+1, c+2, d+3)$ so $$ 1 \leq a^{\ast} < b^{\ast} < c^{\ast} < d^{\ast} \leq 9 $$ and the answer is naturally $9\choose{4}$.
I'm wondering how this statement works and why it is true, is it equivalent to the stars and bars argument?
Thank you.
Lets look at one example. How about $1,3,3,5$?
In stars and bars, this might be represented as:
where the $5$ bars make $6$ "bins" and the number of stars in each bin represent how many times that die-face occurs.
In the new method, we have $(a,b,c,d) = (1,3,3,5) \implies (a^*, b^*, c^*, d^*) = (1, 4, 5, 8)$ and these are exactly the locations of the stars (if the locations are numbered $1$ through $9$).
I know an example is not a proof, but there's nothing special about $1,3,3,5$. The methods are equivalent - under the assumption that the dice faces are consecutive integers. Under that assumption, I do like the new method, as it can be explained very easily. However if your dice have faces which are not consecutive (e.g. $1,2,3,7,8,25$) then the new method would become so messy it's not worth the trouble.