Books on a shelf!

I will be denoting all chosen books as $C$ and not chosen as $N$ .

So the books can be listed as $C$ $C$ $C$ $C$ $C$ $N$ $N$ $N$ $N$ $N$ $N$ $N$

We don't want any $C$ together. There should be atleast one $N$ between $2$ $C$ 's.

$\_\_$ $N$ $\_\_$ $N$ $\_\_$ $N$ $\_\_$ $N$ $\_\_$ $N$ $\_\_$ $N$ $\_\_$ $N$ $\_\_$

By placing $C$ 's in any of the blanks, the required condition can be satisfied.

The number of ways to place $5$ $C$ 's in $8$ boxes is

$=$ $8 \choose 5$

$=\frac{8 \times 7 \times 6 \times 5 \times 4}{5 \times 4 \times 3 \times 2 \times 1}$

$= \boxed{56}$

Let $a_i$ be the position of the $i$ th book you choose, so $1 \leq a_1 < a_2 < a_3 < a_4 < a_5 \leq 12.$ No two chosen books stand next to each other, so the sequence $a_1,\quad a_2-1,\quad a_3-2,\quad a_4-3,\quad a_5-4$ is strictly increasing. Now we know that $1 \leq a_1 < a_2-1 < a_3-2 < a_4-3 < a_5-4 \leq 12-4 = 8,$ so the problem has reduced to finding the number of ways to pick $5$ elements from a set of $8$ elements, hence the answer is ${8 \choose 5} = \boxed{56}.$

We can ask this problem another way.

We have 5 books, how many ways are there of putting 7 books ( $7+5=12$ ) in between or next to them are there, such that the original books are not adjacent? Obviously, the first four books we place have to go in between the original books. That leaves us with 3 books which we can place as we please. Here is a diagram to clarify the situation: the original books are denoted $O$ and the new books $N$ :

$O N O N O N O N O$

Here, the potential positions for a new book are denoted by an "?" :

$? O ? N ? O ? N ? O ? N ? O ? N ? O ?$

Placing a book either to the left or to the right of an N is the same thing, so we have 6 potential positions and 3 books to place. We're now basically asking how many ways are there to place k balls in n boxes. The answer to this is $\binom{k+n-1}{k}$ Here $k=3$ and $n=6$ :

$\binom{3+6-1}{3} = \binom{8}{3} = \frac{8*6*5}{1*2*3} = 56$

Notice that there is an additional condition of having

no two of the chosen books stand next to each other.

Observe that there will be at most $4$ gaps without any other books between the $5$ chosen books if they are all next to each other. Hence, first remove $4$ books from the shelf and we're left with $12-4 = 8$ books.

From the remaining $8$ books, we can choose any $5$ to be our chosen books, without the added condition when there is $12$ books. This is because we can simply add in each of the $4$ books taken out into the space between each pair of books chosen such that no books will be next to each other.

Hence, the number of ways is $8\choose 5$ $= \boxed{56}$

We will use the rest 7 books as a separators for the 5 books, the 7 separators makes 8 available places to put a book there.

so the solution is $\binom{8}{5}=\boxed{56}$