Algebra - Permutation

1057

Ketan Chandak ·2011-10-06 22:29:02

lets assume cases here:
1)E _ _ _ _
in dis case other four can be arranged in 4!=24 ways..

2)_ E _ _ _
in dis case the first letter has 3 options(cant be U),3rd letter has 3 options,4th has 2 options nd de last 1ne has 1 option
total no of ways=3*3*2*1=18 ways

3) _ _ E _ _
in dis case de first letter has 3 options nd 2nd and 4th letter has 2 options nd de last 1ne has 1 option
no of ways = 3*2*2*1=12

4)_ _ _ E _
in dis case de last letter has to be u nd first three can be arranged in 3! ways.....=6 ways....
the last letter cant be E

total ways=24+18+12+6=60

actually another gud way of thinking dis is either de order will be E before U or E after U
so total ways will be 5!2 for each case....

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xYz ·2011-10-07 04:15:54

replace all vowels by x ..
so new word is

VxNxS

no.of ways of permutation is 5!2!

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sagnik007 bhattacharya ·2011-10-07 13:28:01

to mantain the correct order "u" must come after "e"
So when e in frst place the rest can re arrange in 4! ways.
when in 2nd place frst place can be filld in 3 ways nd in each case d rest can re arrange in 3! ways. i.e 3*3!=18 ways..
similarly when in nxt place they cn arrange in 3p2 i.e 3! ways multiplied by 2 ways i.e 12 ways.
similarly in nxt case they ca rearrange in 3! ways
total no of ways =24+18+12+6= 60 ways