Consider $5$ dice with six sides, three of which are labeled $1$ and three of which are labeled $2$. How many ways are there to get a sum of $9$?

Solution 1:

$(3x+3x^2)^5=(3x(1+x))^5$, we know that $(a\times b)^k =(a^k \times b^k)$ , then $(3x(1+x))^5=(3x)^5(1+x)^5$ , so we have exponenetial $5$ form $(3x)^5$ , then find the coefficient of $x^4$ in the expansion $(1+x)^5$ to reach $x^9$ using binomial expansion such that $(1+x)^5= \binom{5}{n}(1)^{5-n}(x)^n$ , so we need the case where $n=4$.

Then , $3^5 \times C(5,4) =243 \times 5 =1215$

Solution 2:

Im not really sure but i think that, for each dice there are 3 ways to get 1 and 3 ways to get 2. The only possible way to get 9 as a sum is (2,2,2,2,1) and all permutations wich are 5 if we consider that the 5dice are distinct. Now, since every single dice has 3 ways to get either 2 or 1 and we roll all the dice together we have $3^5$ ways. So by the addittive and multiplicative law, the answer is $5\cdot 3^5$.