Question

In how many ways can $13$ question papers be arranged so that the best and the worst are never come together?

• A. $12!\times 11$
• B. $11\times 12!$
• C. $12{!}^2$
• D. none of these

Answer 1

The correct option is A $12!\times 11$

Given that there are $13$ papers $\xi$ the best $\xi$ worst should not be together.

Not be together $=$ Total $-$ Together.

$\Rightarrow$ Total Permutations/ arrangements $=13!$

$\Rightarrow$ Best $\xi$ worst are together $:$ The $2$ papers are to be considered as one entity $\xi$ arranged with others, this can be done in $(13-2)+1!=12!$

$\Rightarrow$ But the papers positions can be interchanged $=12!\times 2$

$\Rightarrow$ Req $=13!-2\times 12!=12!(13-2)=12!\times 11$

Hence, the answer is $12!\times 11.$