All combinations of three different dices of side $4, 5$ and $6$.

Question

You have three dice each with a different number of sides, and numbers ranging from $1$ to the number of sides.

$d_1 = 1 \sim 4$

$d_2 = 1 \sim 5$

$d_3 = 1 \sim 6$

How many possible combinations are out there rolling these three dice? ($[1,1,2]$ will be the same as $[1,2,1]$ and $[2,1,1]$)

I tried to hard figuring this out with combinations but I can't get my head around it.

Answer 1

First, let's figure out how many combinations we can have with three $6$-sided dice. It's the same as counting the number of solutions of the equation $$x_1+x_2+x_3+x_4+x_5+x_6=3$$ in nonnegative integers; by the so-called "stars and bars" formula, the answer is $\binom83=56$.

Not all of these $56$ combinations are possible with your actual set of dice, some of them have too many $5$'s and/or $6$'s; but the number of impossible combinations is small enough to count by hand.

Three sixes: $[6,6,6]$
Two sixes: $[6,6,1]$, $[6,6,2]$, $[6,6,3]$, $[6,6,4]$, $[6,6,5]$
One six and two fives: $[6,5,5]$
Three fives: $[5,5,5]$

That's $8$ impossible combinations, so the answer to your question is $56-8=48$.

Answer 2

Let list all possible combinations together with their cardinality ($xyz$ mean numbers less than or equal to 4): $$\begin{align} &[x56]:&4\\ &[xy6]:&4+\frac{12}2=10\\ &[x55]:&4\\ &[xy5]+[x5y]:&4+\frac{12}2=10\\ &[xyz]:& 4+\frac{36}3+\frac{24}{6}=20\\ & \text{all}:&48 \end{align} $$

Explanation:

$4$ stays for possible combinations with $xy$ or $xyz$ being equal.

$12=4\cdot3$ stays for possible combinations with distinct $xy$. It should be divided by $2=\frac{2!}{1!1!}$ to avoid double counting.

$36=\binom31 4\cdot3 $ stays for possible combinations with two of $xyz$ being equal and distinct from the third one. It should be divided by $3=\frac{3!}{1!2!}$ to avoid double counting.

$24=4\cdot3\cdot2 $ stays for possible combinations with all three of $xyz$ being distinct. It should be divided by $6=\frac{3!}{1!1!1!}$ to avoid double counting.