Permutations where two of the same color cannot be adjacent.

Question

A child has 16 crayons of which 10 are blue and 6 are red. Crayons of the same color are indistinguishable from each other. In how many ways can she arrange the crayons in a row such that no two red crayons are next to each other?

Answer 1

Take five blue and six red crayons. Put them in a row any way. Add an extra blue crayon between the first red and second red, put one between the second and third reds etc. You insert five blue crayons overall. Show by this method you get all allowable arrangements of ten blue and six red crayons.

Answer 2

If you line up all $10$ blue crayons, there will be $11$ 'gaps' in which we can insert a red crayon. Since we can't insert two red crayons into the same gap, we must choose exactly $6$ of these gaps to insert red crayons so the answer we're looking for is $\binom{11}{6} = 2310$