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In how many ways can $10$ adults and $5$ children stand in a line so that no two children are next to each other?

My solution: $_A_A_A_A_A_A_A_A_A_A_$

So I'm trying to follow this example:Combination and Permutation, and write:

$10!/2!\cdot\binom{11}{5}\cdot 5!/2!$

Is this correct?

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The $10$ adults can stand in $10!$ ways. The $5$ children have $11$ places to stand so that there is an adult next to them on either side. We can place the $5$ children in ${11\choose 5}$ ways and then order the $5$ children in $5!$ ways. Thus by the Multiplication Principle there are ${11\choose 5}\cdot 5!\cdot 10!$ ways to place the $10$ adults and $5$ children so that no two children are standing next to each other.

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