Question

In how many ways can $9$ books be arranged on a shelf if $2$ of them must be kept together?

Answer 1

Hint: We will use combination to find how many ways we can arrange the books. We will consider the two books that should be kept together as one. Then we will find how many ways we can arrange the books.

Complete step by step solution:
Let us consider the $9$ books to be arranged on a shelf. We are given the condition that two of the books must be kept together.
Let us consider the two books as one. Then, we have $8$ books to be arranged.
We know that we can find the number of ways the $8$ books can be arranged using combination.
We know that the first book has $8$ choices after which $7$ choices are remaining for the next book. So, we can say that there are $6$ choices for the third book, then $5$ choices for the fourth book. We can find $4$ choices for the fifth book and $3$ choices for the sixth book. Now, for the remaining two books, we can find $2$ choices and $1$ choice respectively.
So, we will get $8!=1\times 2\times 3\times 4\times 5\times 6\times 7\times 8=40320.$
This implies that we can arrange $8$ books in $40320$ ways.
We know that the two books that are kept together can be arranged in two ways. So, we will get the total number of ways we can arrange the $9$ books.
Hence, we can arrange the $9$ on a shelf in $40320\times 2=80640$ ways.

Note: We should not forget that the two books that are kept together can also be arranged in two ways. So, for each of the arrangements, the eight books can be arranged in $40320$ ways. Therefore, we need to multiply $40320$ with $2$ to get the possible number of arrangements of $9$ books.