namurad wrote:

rpmodi wrote:

itsme291 wrote:

Cant be 3+3+3. Try doing it.

A B C D

a b c d

write the lowercase alphabets below the upper case so that no pair has the same alphabets.

even with that method my answer is 9 , am I missin something

A B C D

b c d a

b a d c

b d a c

c a d b

c d b a

c d a b

d a b c

d c b a

d c a b

Yes Mr. Modi..u r bang on target. Could u please explain the approach of reaching the ans without listing down all options? This listing down options business is too time consuming.

diffrent approaches here

1) First as I have listed all the combination , just list the numbers for one letter being the wrong envelop and then pattern repeats so you can simply add and that's how I got 3+3+3

For e.g in the above example , I just have to jot down different combinations for letter b in envelop A , since I got 3 combinations , pattern will repeat for c and d . (obviously you have to rule out all the combinations of letter a in envelop A )

A B C D

b c d a

b a d c

b d a c

Itsme's answer would have been correct if it were asked what are the different ways that atleast one letter is in correct envelop .