Bookworm's Choice

We called $R$ the books (they are $5$ ) that we choose and $N$ the books tha t we don't choose (they are $7$

If we put in the row $N$ books, in order to not put two adjacent books $R$ , we put them in $P_{5}^{8}$ (there are $8$ places between the $7$ books $N$ , and we have to put 5 books $R$ ). Then we have to divide by $5!$ (the $R$ books are equal)

$T = \frac {P_{8}^{5}}{5!} \rightarrow \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4}{5!} = 56$

let there be a0 number of books on the left of 1st book and a(i) denote the number of books between (i)th and (i+1)th selected book for i =1,2,3,4 and a5 be the number of books on the right of last ( 5th) selected book.

so we have a0,a5>=0 and a (i)>=1

and a0+a1+a2+a3+a4+a5=7 number of ways is same as the number of solutions of the above which is found from multinomial to be 8C3.