# Counting Dice Outcomes

(Multinomials)

### The formula...

\tiny \dfrac{n!}{\text{Intrapile swappability}} \times \dfrac{F!}{\text{Interpile swappability}}

One multinomial to assign trials to piles

One multinomial to assign faces (ranks) to pile sizes

n = number of trials (rolls)

F = number of faces (sides) on the die

Problem 1

If we roll a 6-sided die 5 times, in how many ways can we get 3 of a kind (3OAK) and 2 of a kind (2OAK)?

(Click the Down Arrow to see the solution)

\tiny \dfrac{n!}{\text{Intrapile swappability}} \times \dfrac{F!}{\text{Interpile swappability}}
\tiny \color{blue} \dfrac{5!}{\text{3!2!0!0!0!0!}} \times \dfrac{6!}{\text{1!1!4!}} = 10 \times 30 = 300

3OAK     2OAK    0OAK     0OAK     0OAK     0OAK

Intrapile         3             2           0            0             0             0

Interpile         1             1                                  4                           of size 3        of size 2                                      of size 0

(Click Right Arrow for the next problem)

There are F=6 piles--one for each side of the die.

F=6 piles  →

1 pile of size 3, 1 pile of size 2, 4 piles of size 0

Problem 2

Using an 8-sided die, in how may ways can we get 3OAK, 3OAK, and 1OAK?

(Click Down Arrow for the solution)

\tiny \color{blue} \dfrac{7!}{3!3!1!0!0!0!0!0!} \times \dfrac{8!}{2!1!5!} = 140 \times 168 = \text{23,520}
\tiny \dfrac{n!}{\text{Intrapile swappability}} \times \dfrac{F!}{\text{Interpile swappability}}

(Click Right Arrow for the next problem)

F=8 piles  →

2 piles of size 3, 1 of size 1, 5 of size 0

Problem 3

Using a die with 5 sides, in how many ways can we get 2OAK, 2OAK, and 2OAK?

(Click Down Arrow for the solution)

\tiny \dfrac{n!}{\text{Intrapile swappability}} \times \dfrac{F!}{\text{Interpile swappability}}
\tiny \color{blue} \dfrac{6!}{\text{2!2!2!0!0!}} \times \dfrac{5!}{\text{3!2!}} = 90 \times 10 = 900

F=5 piles  →

3 piles of size 2, 2 piles of size 0

Fin

By smilinjoe

• 596