1.In how many ways can 10 exam papers be arranged so that the best and worst papers never come together .?
1.In how many ways can 10 exam papers be arranged so that the best and worst papers never come together .?
total number of ways in which 10 examination papers be arranged = 10!
now total number of ways in which examination papers are arranged such that best and worst paper always come together =9!*2!
so number of ways in which they never come togerther = 10! -9!*2!
=8*9!
here's the orignal one
http://www.pagalguy.com/discussions/official-quant-thread-for-cat-2011-part-1-25060879
so post here.
mods do the needful .:grin:
total number of ways in which 10 examination papers be arranged = 10!
now total number of ways in which examination papers are arranged such that best and worst paper always come together =9!*2!
so number of ways in which they never come togerther = 10! -9!*2!
=8*9!
thanks but my question is when best and worst papers come together why are we taking them as 1 and subtracting 1 paper from 10 papers to get 9 and their arrangement as 9 !
i mean best paper and and worst paper cant be same as 1 .. it shud be 2 diferent papers so
whi is not 10 - 2 = 8
and arrange of 8!
final result 10!-8!x2!
thanks but my question is when best and worst papers come together why are we taking them as 1 and subtracting 1 paper from 10 papers to get 9 and their arrangement as 9 !
i mean best paper and and worst paper cant be same as 1 .. it shud be 2 diferent papers so
whi is not 10 - 2 = 8
and arrange of 8!
final result 10!-8!x2!
consider the worst and best paper as one group......now what's remaining, are 8 other papers...so total of 9 papers.....8 other papers and one paper group of worst and best papers...total ways in which these papers can be permuted is 9!*2!...
2! coz worst and best paper can permute b/w themselves in 2! ways......so
it will be 10!-9!*2!......