Question

There are $n+1$ identical balls on which the terms ${}^n{C}_0{,}^n{C}_1{,}^n{C}_2.........{.}^n{C}_n$ are written, each ball carrying a different term (not necessarily the value of the term). The total number of ways in which the balls can be arranged in a row so that the values of numbers on the first and the last are equal, is

• A. $n!$ when n is even
• B. $(n^2-1)\cdot (n-2)!$ when n is odd
• C. $\frac{1}{2}\cdot n$! when n is even
• D. $\frac{1}{2}(n+1)\cdot (n-1)!$ when n is odd

Answer 1

Answer: (A), (B)

The correct options are
A $n!$ when n is even
B $(n^2-1)\cdot (n-2)!$ when n is odd

We take two cases,
(i) $n$ is even
The number of pairs of equal numbers is $\frac{n}{2}$ since ${}^n{C}_r{=}^n{C}_{n-r}$
We need to multiply by $2!$ since the first and the last can be interchanged.
Now, the remaining $n-1$ balls can be arranged among themselves in $(n-1)!$ ways.
Hence the total number of ways become $\frac{n}{2}\times 2!\times (n-1)!=n!$
(ii) $n$ is odd
The number of pairs of equal numbers is $\frac{n+1}{2}$ since ${}^n{C}_r{=}^n{C}_{n-r}$
We need to multiply by $2!$ since the first and the last can be interchanged.
Now, the remaining $n-1$ balls can be arranged among themselves in $(n-1)!$ ways.
Hence the total number of ways become $\frac{n+1}{2}\times 2!\times (n-1)!=(n+1)\times (n-1)!=(n^2-1)\times (n-2)!$