In how many ways can 10 examination papers be arranged so that the best and the worst papers never come together?

The question asks for the number of ways to arrange 10 examination papers so that the best and the worst papers never come together.

This type of problem, where certain items "never come together", has a common strategy:
Total Arrangements - Arrangements Where They ARE Together = Arrangements Where They Are NEVER Together

Let's break it down step-by-step:

Step 1: Calculate the Total Number of Ways to Arrange all 10 Papers.
If we have 10 different papers, and we want to arrange them in a line, the number of ways is given by 10 factorial (10!).
10! means 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1.
So, Total arrangements = 10!

Step 2: Calculate the Number of Ways where the Best and Worst Papers ARE Together.
To make sure the best and worst papers are always together, we can think of them as a single "block" or "unit".
Imagine tying the best paper and the worst paper together. Now, instead of 10 individual papers, you have:
(Paper 1), (Paper 2), ..., (Paper 8), and (Best & Worst Paper Block).
This means you now have 9 items to arrange (8 individual papers + 1 block).
These 9 items can be arranged in 9 factorial (9!) ways.

BUT, inside the "Best & Worst Paper Block", the best paper and the worst paper can swap their positions! They can be arranged in 2 factorial (2!) ways:
(Best, Worst) or (Worst, Best)

So, the total number of arrangements where the best and worst papers are together is: 9! × 2!

Step 3: Calculate the Number of Ways where the Best and Worst Papers NEVER Come Together.
Now we use our strategy:
Never Together = Total Arrangements - Arrangements Where They ARE Together
Never Together = 10! - (9! × 2!)

Step 4: Simplify the Expression.
We know that 10! can also be written as 10 × 9!
And 2! is 2 × 1 = 2.

So, our equation becomes:
Never Together = (10 × 9!) - (9! × 2)
You can see that 9! is common in both parts. Let's factor it out:
Never Together = 9! × (10 - 2)
Never Together = 9! × 8
Or, simply, 8 × 9!

This matches Option A.

Alternative way
No. of ways in which 10 paper can arranged is 10! Ways.
When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers.
These 9 papers can be arranged in 9! Ways.
And two papers can be arranged themselves in 2! Ways.
No. of arrangement when best and worst paper do not come together,
= 10! - 9! × 2!
= 9!(10 - 2)
= 8 × 9!