in how many different ways can 8 examination papers be arranged in a row, so that the best and the worst papers may never come together?

The number of ways is 30240.

Quick Scoop: Arranging 8 Examination Papers

Final Answer

In how many different ways can 8 examination papers be arranged in a row, so that the best and the worst papers may never come together?
→ 30240 ways.

Step-by-step idea (short and clear)

Total arrangements without any restriction
There are 8 distinct papers, so total arrangements:

8!=403208!=403208!=40320

Arrangements where best and worst are together
- Treat “best + worst” as one block.
- Then we have 7 units (this block + 6 other papers).
- These 7 units can be arranged in 7!7!7! ways.

 * Inside the block, “best, worst” can be ordered in 2!2!2! ways (BW or WB).

So, arrangements with best and worst together:

7!×2!=5040×2=100807!\times 2!=5040\times 2=100807!×2!=5040×2=10080

Arrangements where they are never together

8!−(7!×2!)=40320−10080=302408!-(7!\times 2!)=40320-10080=302408!−(7!×2!)=40320−10080=30240

Tiny “story” way to remember it

Think of 8 papers marching in a line.
First count all possible marching orders.
Then count the “bad” ones where the best and worst are handcuffed together as a single pair.
Subtract those “handcuffed” orders from the total, and you’re left with only the good ones — where they never stand side by side.

HTML table summary

html

<table>
  <tr>
    <th>Quantity</th>
    <th>Expression</th>
    <th>Value</th>
  </tr>
  <tr>
    <td>Total arrangements of 8 papers</td>
    <td>8!</td>
    <td>40320</td>
  </tr>
  <tr>
    <td>Arrangements with best & worst together</td>
    <td>7! × 2!</td>
    <td>10080</td>
  </tr>
  <tr>
    <td>Arrangements where they never come together</td>
    <td>8! − (7! × 2!)</td>
    <td><strong>30240</strong></td>
  </tr>
</table>

TL;DR:
Number of arrangements where the best and worst papers never come together = 30240.

Information gathered from public forums or data available on the internet and portrayed here.