in how many ways can:

$10$ adults, $20$ boys and $40$ girls be arranged in a row such that between every $2$ adults there are exactly $2$ boys and $4$ girls?

i figured the adults go in places $1,7,13...$ and to choose their positions there are $10!$ combinations, how do i find the sum of all the arrangements including the boys and girls?

  • $\begingroup$ What happens at the ends of the row? Can't you have CCP...PCCCC? (P=parent, C=child) $\endgroup$ – Ragnar Dec 27 '13 at 22:24

Let's just assume that there is a parent at the left end of the row and $6$ childs at the right end. For the parents, there are indeed $10!$ possibilities. Next, we determine the order in which the boys appear in the row. There are $20!$ possibilities for that. For the girls, there are $40!$ possibilities. Now we select the first two boys and the first four girls for the first group of children. We have to select in which two of the six places the boys stand. This is possible in $6\choose 2$ ways. Because there are $10$ groups of children, we get $\binom 62^{10}$ ways of doing this. Al together, we get $$ 10!20!40!\binom62^{10} $$ ways of ordering the people.

If there may be a child at the front of the row, we can rotate the found rows $1$, $2$, $3$, $4$, $5$ or $6$ places, so we get $$ 7\cdot10!20!40!\binom62^{10} $$


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