Permutations

Let us make blocks first of 5, 3, 2

first block of 5 starts at position A

Block of 3 will start at position A+5+x

Block of 2 will start at A+5+x+3+y

such that
A+5+x+3+y+2 <= 15

a+x+y<=5

so we have another z

a+x+y+z=5

divide 5 in 4 parts we need 3 dividers

hence 8!/{3! . 5!} blocks can be arranged in 3! ways
and the blocks can be given to A, B or C in 3! ways..

I dont know why i am not convinced..but i think some part of this solution is a creativity of my mind :(