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

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Hint: use the concept of permutation and combination to find the total number of arrangements and then subtract it with arrangement when best and worst paper are together to get the arrangements in which best and worst paper are not together.
Complete step-by-step answer:
No. of ways in which ways can 10 examination papers be arranged so that the best and worst papers never come together = Total arrangements – Best and worst together
Total arrangements = n!
                     = 10!
Best and worst together = 9!$ \times $2 (As we assumed best and worst as a group so total papers became 9 so used 9! and both best and worst can interchange their positions so multiplied with 2)
Required number of ways = 10! - 9!$ \times $2
                   = (10$ \times $9!) - (9! $ \times $ 2)
                   = 9! (10-2)
                   = 362880$ \times $8
                   =2903040.
Note: Use the concepts of permutation and combination to find the total number of arrangements and then subtract it with the non-favourable arrangement to get the total number of favourable arrangements. Students often forget to multiply 9! by 2 so please take care of that.

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