Fastidious Librarian

The same problem can be interpret into the following :

The bookshelf has 6 distinct books and you have 4 books to put in between them such that there is no books among the 4 is consecutive. How many ways can you do that?

Since there are 7 gaps and 4 books to put in to, there are $7 \choose 4$ ways

Let the books be $A,B,C,D,E$ , then their placement is as such: $- \ A \ - \ B \ - \ C \ - \ D \ - \ E -$

where each bar between $A$ and $E$ denote a nonzero number of books.

Let these bars be denoted as $v,w,x,y,z$ for integers, $v,z \geq 0, w,x,y \geq 1$

Which satisfy the equation $v+w+x+y+z = 10 -4=6$

Let $w' = w - 1, x' = x-1,y' = y-1$ , then we need to calculate the non negative solution for $v+w'+x'+y'+z = 3$ which is simply ${ 5+3-1 \choose 3 } = \boxed{35}$

Let the books be numbered from $1$ to $10$ . There are $9$ pairs of consecutive books possible $(\{1, 2\}, \{2, 3\}, ..., \{9, 10\})$ .

Total number of ways to choose $4$ books $= 10C_{4}$

Total number of ways to choose $4$ books such that it has at least one consecutive pair $=$ $9C_{1} \times \frac{8 \times 7}{2!}$

Explanation : For a given consecutive book pair $\{a,a+1\}$ , the $3^{rd}$ book can be chosen in $8$ and the $4^{th}$ book can be chosen in $7$ ways. And there are total $9C_{1}$ ways to choose a consecutive pair.

Total number of ways to choose $4$ books such that it has at least two consecutive pairs $=$ $9C_{2} + 8C_{1} \times 6$

Explanation : There are two ways of choosing two consecutive pairs, viz. non-overlapping and overlapping case.

non-overlapping case :

e.g. $\{1, 2, 7, 8\}$ has two non-overlapping consecutive pairs viz. $\{1, 2\}$ and $\{7, 8\}$ . There are $9C_{2}$ ways of choosing two non-overlapping consecutive pairs.

overlapping case :

e.g. $\{1, 2, 3, 8\}$ has $2$ overlapping consecutive pairs viz. $\{1, 2\}$ and $\{2, 3\}$ . There are $8$ overlapping consecutive pairs possible which are $\{1, 2, 3\}, \{2, 3, 4\}, ..., \{8, 9, 10\}$ . For a given overlapping consecutive pair, there are $7$ ways to choose the remaining $4^{th}$ book. However, one choice of the $4^{th}$ book makes the two overlapping pair case as two non-overlapping consecutive pairs case as well. Hence this $1$ choice has already been counted in the non-overlapping case discussed above. e.g. for the overlapping case $\{1, 2, 3, -\}$ , choosing $4$ as the $4^{th}$ book introduces non-overlapping case as well i.e. $\{1, 2\}$ and $\{3, 4\}$ . Hence there are effectively $6$ ways to choose the $4^{th}$ book.

Total number of ways to choose $4$ books such that it has at least three consecutive pairs $=$ $7C_{1}$

This is possible only when all the four books are consecutive, e.g. $\{1, 2, 3, 4\}, \{2, 3, 4, 5\}, ..., \{7, 8, 9, 10\}$ .

Using principle of exclusion-inclusion:

Total ways of choosing books such that no consecutive books are chosen $=$ $10C_{4} -(9C_{1} \times \frac{8 \times 7}{2!} - 9C_{2} - 8C_{1} \times 6 + 7C_{1})$ $=210 - (9 \times 28 - 36 - 48 + 7) = 210 -175 = 35$