Question

The number of ways in which ten candidates ${A_1},{A_2},.{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{{\rm{A}}_{{\rm{10}}}}$ can be ranked such that ${A_1}$ is always above ${A_{10}}$ is
(a) $5!$
(b) $2\left( {5!} \right)$
(c) $10!$
(d) $\dfrac{1}{2}\left( {10!} \right)$

Answer 1

Hint: We will first select 2 positions out of 10 such that ${A_1}$ is always above ${A_{10}}$ and then remaining candidates are arranged to the remaining position. We will use the concept that selecting r thing out of n things is equal to ${}^n{C_r}$ and arrangement of n things is equal to \[{\rm{n}}!\].

Complete step-by-step solution -
We have been asked to find the number of ways in which ten candidates ${A_1},{A_2},.{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{{\rm{A}}_{{\rm{10}}}}$ can be ranked such that ${A_1}$ is always above ${A_{10}}$
We have ten positions. So, firstly we will select 2 positions out of 10 for ${A_1}{\rm{ and }}{{\rm{A}}_{{\rm{10}}}}$ such that ${A_1}$ is always above ${A_{10}}$
Some possible ways are,

${A_1}$

${A_{10}}$

${A_1}$

${A_{10}}$

So, total number of ways $ = {}^{10}{C_2}$
Since, the 2 positions is already filled by ${A_1}{\rm{ and }}{{\rm{A}}_{{\rm{10}}}}$ , we have remaining (10-2) i.e., 8 position, which we have to arrange by the remaining candidates.
Remaining candidates after arranging,
${A_1}{\rm{ and }}{{\rm{A}}_{{\rm{10}}}} = 10 - 2{\rm{ = 8}}$
So, arrangement of 8 candidates to 8 positions \[ = {\rm{8}}!\]
Required number of ways $ = {}^{10}{C_2} \times 8!$
$\begin{array}{l} \Rightarrow \dfrac{{10!}}{{8! \times 2!}} \times 8!\\ \Rightarrow \dfrac{{10!}}{{2!}}\end{array}$
Therefore, the correct option is (d).

Note: The common silly mistake that we do while choosing the option is that by mistake we write $\dfrac{{10!}}{{2!}} = 5!$ and we choose the option (a). Thus, we get the incorrect answer. So, be careful while choosing the option. We can also solve this question by considering ${A_1}{\rm{ and }}{{\rm{A}}_{{\rm{10}}}}$ as similar elements. Since their order of arrangement is not important, we will have to arrange 10 things out of which 2 are the same and we get $\dfrac{{10!}}{{2!}}$.