Number of way of choosing set of 5 books from 15

Question

Consider a bookshelf with $15$ books placed in sequential manner.

In how many ways one can choose a set of $5$ books from this shelf so that no $2$ books in this set should be adjacently positioned at the time of picking ?

I tried this problem but was not able to find any way of interpreting it with combination and used brute force which in result gave wrong number.

Answer 1

To be found is the number of sums: $$a_1+a_2+a_3+a_4+a_5+a_6=10$$ where $a_1$ and $a_6$ are nonnegative integers and $a_2,a_3,a_4,a_5$ are positive integers.

Here $a_1$ stands for the number of books utmost right of the selected books, $a_6$ for the number of books utmost right of the selected books, and the other $a_i$ correspond with the number of books that divide two selected books (hence prevent them to be adjacent).

Setting $b_1=a_1$, $b_6=a_6$ and $b_i=a_i-1$ for $i=2,3,4,5$ that comes to the same as finding the number of sums: $$b_1+b_2+b_3+b_4+b_5+b_6=6$$ where $b_i$ is a nonnegative integer for $i=1,2,3,4,5,6$.

This can be solved with stars and bars.

Give it a try yourself now, and eventually come back if you get stuck.

Answer 2

The answer will be the same as # of ways of inserting each of the $5$ chosen books in gaps between the $10$ books left.

Don't forget the ends !