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?

1 hour ago 2

Let's analyze the problem step-by-step.

Problem:

You have 8 examination papers, including the best and the worst papers. You want to arrange all 8 papers in a row such that the best and the worst papers never come together.

Step 1: Total number of ways to arrange 8 papers

Since all 8 papers are distinct, the total number of arrangements is:

8!=40,3208!=40,3208!=40,320

Step 2: Number of arrangements where the best and worst papers **are

together**

Consider the best and worst papers as a single combined unit. This combined unit can be arranged internally in 2 ways: (Best, Worst) or (Worst, Best). Now, instead of 8 papers, we have:

1 combined unit (best + worst)
6 other papers

So, total units to arrange = 7 Number of ways to arrange these 7 units:

7!=5,0407!=5,0407!=5,040

Since the combined unit can be arranged internally in 2 ways:

Number of arrangements with best and worst together=2×7!=2×5,040=10,080\text{Number of arrangements with best and worst together}=2\times 7!=2\times 5,040=10,080Number of arrangements with best and worst together=2×7!=2×5,040=10,080

Step 3: Number of arrangements where best and worst are not together

Total arrangements−Arrangements with best and worst together=8!−2×7!=40,320−10,080=30,240\text{Total arrangements}-\text{Arrangements with best and worst together}=8!-2\times 7!=40,320-10,080=30,240Total arrangements−Arrangements with best and worst together=8!−2×7!=40,320−10,080=30,240