No. of ways a specific outcome can occur when 12 die are thrown simultaneously

Question

Twelve normal and fair dice are thrown once. What is the total number of ways in which each of the values $1,2,3,4,5,6$ occur exactly twice? e.g. $(1,1,2,2,3,3,4,4,5,5,6,6)$

My attempt:

The above problem would be similar to the to the number of arrangements of $12$ digits $(1,6)$ out of which each digit is repeated twice… and hence the number of ways the above outcome can occur will be equal to $\frac{12!}{6!}$.

But it seems that my answer is wrong; the answer key reports the answer to be $\frac{12!}{2^6}$.

Is my approach to the problem incorrect? If so, then could you please suggest a way to solve the problem?

Answer 1

The number of ways it can occur is going to be the Multinomial Coefficient $\binom{12}{2,2,2,2,2,2} = \frac{12!}{2!2!2!2!2!2!}$

A way in which you can see this without using multiniomial coefficients directly (but is equivalent to, it is afterall how they were developed) is to approach via multiplication principle with the following steps:

• Pick the location of the two $1$'s: $\binom{12}{2}$ ways

• From the remaining locations, pick the location of the two $2$'s: $\binom{10}{2}$ ways

• From the remaining locations, pick the location of the two $3$'s: $\binom{8}{2}$ ways

  $\vdots$

• From the remaining locations pick the location of the two $6$'s: $\binom{2}{2}$ ways

multiplying these together gives the same result.

Answer 2

If we were considering permutations of $12$ distinct numbers, there would be $12!$ permutations,
but each number is repeated twice, so we need to divide by $2!$ for each such occurrence,
thus $\dfrac{12!}{2!2!2!2!2!2!} = \dfrac{12!}{2^6}$