Permutation and combination question

msd_2008 wrote:Guys,

Can someone help me solve the following question ?

Q. There are four letters and four directed envelopes. What is the total number of ways in which the letters can be put into evelopes so that each letter is in the wrong envelope?

A.243 B.16 C.9 D.64 E.None of these

The OA is A

Regards
MSD

ways of arrangement: 4!=4*3*2*1=24 ways
(1 letter=4 envs, 2nd letter 3 available envs, 3rd letter 2 available envs)
only one way of arranging the letters in correct envelopes
so possible ways of all wrong arrangement=24-1=23 ways

msd_2008 wrote:Guys,

Can someone help me solve the following question ?

Q. There are four letters and four directed envelopes. What is the total number of ways in which the letters can be put into evelopes so that each letter is in the wrong envelope?

A.243 B.16 C.9 D.64 E.None of these

The OA is A

Regards
MSD

First of all OFFICIAL answer is wrong.

4 letters without any condition can but put

4! = 24 ways so this eliminates, A and D

you can either list the possible ways:

2143
2341
2413
3142
3412
3421
4123
4312
4321

or realize that the first mistake will start with 3 different envolops

and the second one will be another 3 and finally another mistake on the 3rd envolope and the rest will be wrong anyways so

3+3+3 = 9

Hence, C

scoobydooby wrote:

msd_2008 wrote:Guys,

Can someone help me solve the following question ?

Q. There are four letters and four directed envelopes. What is the total number of ways in which the letters can be put into evelopes so that each letter is in the wrong envelope?

A.243 B.16 C.9 D.64 E.None of these

The OA is A

Regards
MSD

ways of arrangement: 4!=4*3*2*1=24 ways
(1 letter=4 envs, 2nd letter 3 available envs, 3rd letter 2 available envs)
only one way of arranging the letters in correct envelopes
so possible ways of all wrong arrangement=24-1=23 ways

each letter is in the wrong envelope!

Your solution is answer to another problem.

logitech wrote:

scoobydooby wrote:

msd_2008 wrote:Guys,

Can someone help me solve the following question ?

Q. There are four letters and four directed envelopes. What is the total number of ways in which the letters can be put into evelopes so that each letter is in the wrong envelope?

A.243 B.16 C.9 D.64 E.None of these

The OA is A

Regards
MSD

ways of arrangement: 4!=4*3*2*1=24 ways
(1 letter=4 envs, 2nd letter 3 available envs, 3rd letter 2 available envs)
only one way of arranging the letters in correct envelopes
so possible ways of all wrong arrangement=24-1=23 ways

each letter is in the wrong envelope!

Your solution is answer to another problem.

if there are total 24 ways of arranging and one is eliminating the possibility of the correct sequence, one is left with all wrong sequences. There is only one way in which letters can be put in correct envelopes

say L1, L2, L3,L4 and E1,E2,E3,E4.
correct match can only be L1E1, L2E2,L3E3,L4E4
only one way this can be done: 1*1*1*1=1

[quote="scoobydooby
say L1, L2, L3,L4 and E1,E2,E3,E4.
correct match can only be L1E1, L2E2,L3E3,L4E4
only one way this can be done: 1*1*1*1=1[/quote]

so that each letter is in the wrong envelope

You are right, of course there is only one way of putting all letters in the correct envelops.

But the question is asking how many ways EACH letter is in the wrong envelops, in your words

L1 will be never at E1, and L2 will be never be at E2..etc...

23 ways includes, all the MISTAKES but question asks something different...

thanks logitech, got your point.

for the first letter to go in wrong env=3 choices (only 1 is correct)
second letter to go in a wrong env= 2 choices (out of 4, is already taken, 1 is the correct, so 2 available wrong choices)

third letter in wrong envelope=1 way (2 taken, 1 correct, 1 wrong)
fourth letter in wrong =1 remaining

so number of arrangments: 3*2*1*1=6 ways

scoobydooby wrote:thanks logitech, got your point.

for the first letter to go in wrong env=3 choices (only 1 is correct)
second letter to go in a wrong env= 2 choices (out of 4, is already taken, 1 is the correct, so 2 available wrong choices)

third letter in wrong envelope=1 way (2 taken, 1 correct, 1 wrong)
fourth letter in wrong =1 remaining

so number of arrangments: 3*2*1*1=6 ways

What if the first letter goes to envelope 2

so you will have 3 choices left for the second envelope ( 1,3 and 4)



2143
2341
2413
3142
3412
3421
4123
4312
4321

Thanks Logitech, for pointing put. i get it now