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

A. 8 × 9!

B. 8 × 8!

C. 7 × 9!

D. 9 × 8!

**Answer: Option A **

__Solution(By Examveda Team)__

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!

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## Comments ( 12 )

Related Questions on Permutation and Combination

A. 3! 4! 8! 4!

B. 3! 8!

C. 4! 4!

D. 8! 4! 4!

A. 7560,60,1680

B. 7890,120,650

C. 7650,200,4444

D. None of these

A. 8 × 9!

B. 8 × 8!

C. 7 × 9!

D. 9 × 8!

👌👌👌👌👌 Tysm for Help.

Special thanks to Kumar chandan

it was really helpful

Waal...you are really smart!

Waal...you are really smart!

10!-9!.2! Solve as below

10.9!- 9!.2! And 9! take common here then we get 9!- (10-2 )=8.9!

10!-9!.2!

We can also write

(10*9!-9!*2!)

And 9! Is taken out, ie. We get

9!(10-2)= 9!*8

See, the question, there have been give that best and worst papers never come together.

In the second step of the solution, best and worst paper have been taken together means,

best + worst = One new type of paper.

And the total examination paper is 10 and when best and worst paper taken together it become (10 -1) i.e. 9.

And these 9 papers can be arranged themselves into 9! ways.

I have a doubt how the 9! came in the answer

Where is your problem? @Gautam and what is the problem that you are facing in question?

Still M not getting it.!!

In the second step,why the best and worst paper have been taken together?

how to solve 10!-9!.2!