//Just had a thought...
This problem can be reframed as:
Map 5 books to 15 positions such that no two books are adjacent.
which can be further reframed as:
Map 5 objects to 15 positions such that no two objects are adjacent.
which can be further reframed as:
No. of bit strings of length 15 such that exactly 5 bits are 1 and bits that are 1 can't be adjacent.
Eg: Bit string 101010101000000 can be interpreted as:
book1 is placed in pos. 1 (assuming we count positions from the lhs of the bit string.)
book2 is placed in pos. 3 and so on....
This problem can be further reframed as:
15 bits = 5 1's + 10 0's
Pos. of 10 0's is fixed: i.e. _0_0_0_0_0_0_0_0_0_0_
In how many ways, 5 1's can be mapped to 11 positions. (blanks seen above.)
So, our problem further reduces to:
In how many ways 5 elements be chosen from 11 elements? //elements = positions.
This can be done in: $\binom{11}{5}$ ways.
