A boy named John threw $3$ dice. In how many ways he can get the sum $10$

combinationscombinatoricspermutations

A boy named John threw 3 dice. In how many ways can get the sum 10?

My Approach:

So there are 3 dice: $a, b, c$. Since they are dice, we can write: $1 \le a,b,c \le 6$. My equation right now is $$a+b+c=10$$ Now because, each die will roll at least a one, I subtract that from the total sum to remove the bound so my new equation has conditions $a,b,c \le 5$ and is $$a+b+c=7$$ Using stars and bars, I got the answer as $36$ but the correct answer is $36-9=27$. I want to know where the $9$ comes from.

Note: Please answer this using stars and bars

EDIT: The approach to my answer stems from this question

EDIT: stars and bars does not account for the upper bound restriction therefore, I needed to remove the $9$ other cases where either of $a, b, c$ exceeded $5$. These include the $6$ combinations of the case $6 + 1 + 0$ and the $3$ for $7+0+0$.

Best Answer

The method of stars and bars counts some additional cases, which are two namely, $(0,1,6)$ and $(0,0,7)$, which can be permuted in $6$ and $3$ ways respectively, accounting for a total of $9$ ways, which need to be removed.

Hope this helps. Ask anything if not clear :)

Related Question