How many distributions of $18$ different objects into $3$ different boxes. How many distributions of $18$ different objects into $3$ different boxes are there with twice as many objects in one box as in the other two combined$?$ 

Soln:

It appears to me as an arrangement problem with repetition of some sort.  As such I did this:  $$\binom{3}{1} \frac{18!}{(2n)! n! n! }$$

i did $\binom{3}{1}$ to account for the fact any of the $3$ boxes could get the double amount.  $\frac{18!}{(2n)! n! n! }$ was done partially because I have seen a formula relating to this sort of problem,  but personally I am trying to understand the REASON behind why this formula works here, that is if my treatment was correct
If you have $18$ items, the big box must have $12$ and the other two must have $6$ between them. There are $18 \choose 12$ ways to select the ones in the big box, $3$ ways to select the big box, now distribute the other $6$.
Out of $18$ we select any of the $12$ objects to be placed into any of the $3$ boxes.This can be done in $3*\binom{18}{12}$ ways.There remain $6$ distinct objects to be distributed into $2$ distinct boxes.This can be done in $64$ ways.    $P(6;0,6)+P(6;1,5)+P(6;2,4)+P(6;3,3)+P(6;4,2)+P(6;5,1)+P(6;6,0)=64...$