Question

n books are arranged on a shelf so that two particular books are not next to each other. There were 480 arrangements altogether. Then the number of books on the shelf is
A. 5
B. 6
C. 10
D. 8

Answer 1

Hint: We first find the complement event where we find the number of arrangements keeping the particular two books together. We subtract that from the unconditional solution to find the required number 480. We need to keep in mind the arrangements for those two books for ourselves.

Complete step-by-step solution:
n books are arranged on a shelf so that two particular books are not next to each other.
We try to find the number of arrangements taking the complement set where we try to find the number of arrangements keeping the particular two books together.
If there are no conditions then the arrangements of the n books will be done in $n!$ ways.
Now we want to keep the two books together.
So, they will be treated as a single one but the arrangement among those two will be done $2!=2$ ways.
Now there are in total $n-2+1=n-1$ books.
The arrangements are $2\times \left( n-1 \right)!$.
Therefore, the number of arrangements where two particular books are not next to each other is \[n!-2\times \left( n-1 \right)!\]. It is given that the number is 480. So, $n!-2\times \left( n-1 \right)!=480$.
We can see from given options that $n=6$ satisfies $n!-2\times \left( n-1 \right)!=480$.
$\begin{align}
  & 6!-2\times \left( 6-1 \right)! \\
 & =720-240 \\
 & =480 \\
\end{align}$
Therefore, the value of n is 6. The correct option is B.

Note: There is no particular method to solve the equation $n!-2\times \left( n-1 \right)!=480$ directly. We have to use the option check to confirm the solution. The method of taking those two books as one is called bundling up. It keeps them fixated together at one place.