You have the following collection of brightly colored balls:

You put them all in a bag, mix them up and them pick them out (without replacement) one by one.

If the probability that the first three you pick will be the same color, the second three will be the same color and the third three you pick will be the same color, is $\frac{1}{a}$ then what is $a$ ?

The answer is 15400.

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The number of ways you can pick out $12$ balls is given by $12!$

The number of ways you can pick them out in groups of $3$ of the same color (as described in the problem) is given by $(3!)^4 \cdot 4!$

The $(3!)^4$ accounts for the number of ways of arranging similar balls within each group. The $4!$ accounts for the number of ways of rearranging the groups themselves.

So, the probability you will pick them out as described above is $\frac{(3!)^4 \cdot 4!}{12!} = \frac{1}{\boxed{15400}}$