Math, asked by Jezneel6003, 1 year ago

10 examination papers are arranged in such a way that the best and worst papers never come tog.The number of arrangement is..??

Answers

Answered by bhagyashreechowdhury
140

If 10 examination papers are arranged in such a way that the best and worst papers never come tog, then the number of arrangement is 8 * 9! or 2903040.

Step-by-step explanation:

There are a total of 10 examination papers.

So, the total no. of arrangements in which, all the 10 examination papers are arranged without any condition = 10! Ways.

Case 1:

Now, let’s see the no. of arrangements that can be done if the best and the worst papers come together.

For this case, we can consider the best and the worst papers to be one, so, the no. of arrangements will be = 10! – 1! = 9! Ways

Also, the best and the worst can be arranged in 2 different ways themselves i.e., [best & worst] or [worst & best].

No. of ways 10 examination papers can be arranged, when the best and the worst papers come together = 2! * 9! Ways  

Case 2:

Thus,  

The no. of ways 10 examination papers can be arranged when the best and worst papers never come together is,

= 10! – [2! * 9!]

= [10*9!] – [2*1*9!]

= 9! [10 - 2]

= 9! * 8

= 9*8*7*6*5*4*3*2*8

= 2903040

Hope this is helpful!!!!!!

Answered by mail2daku
57

Answer:

8*9!

Step-by-step explanation:

there are 10 examination papers = 10!ways

when the best and worst papers come together, regarding 2 papers as 1 paper, we have only 9 papers.

Those papers can be arranged in 2! ways

when both are not together

=10! - 9!* 2!

9! (10-2)

8*9!

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