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PR problem - combinations

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by jayhawk2001 » Sat May 26, 2007 2:18 pm
OA after a few replies.

A secretary types 4 letters and then addresses the 4 corresponding envelopes. In how many ways can the secretary place the letters in the envelopes so that NO letter is placed in its correct envelope?
A. 8
B. 9
C. 10
D. 12
E. 15
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by Ramesh2007 » Sat May 26, 2007 5:51 pm
There will be total of 16 ways including the matching ones. There will be 4 matching ones hence the answer is 12.
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by f2001290 » Sun May 27, 2007 2:40 am
Awaiting the OA. No clue about this problem
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by jayhawk2001 » Sun May 27, 2007 8:15 am
f2001290 wrote:Awaiting the OA. No clue about this problem
I'm eagerly awaiting some alternate approaches to this problem from
folks on this forum. I got this wrong in my PR CAT and wanted to see
if there's a better/easier way to solve this than the long-winded calculate-all
approach that PR has provided.

I'll PM you the OA and the explanation.
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by Prasanna » Sun May 27, 2007 7:30 pm
I tried the long winded way and got 9 as the answer

Assuming there are four letters A,B,C and D to be put into 4 covers 1,2,3 and 4 respectively, the different ways in which the letter can be wrongly covered are

A2,B1,C4,D3
A2,B3,C4,D1
A2,B4,C1,D3

A3,B1,C4,D2
A3,B4,C1,D2
A3,B4,C2,D1

A4,B1,C2,D3
A4,B3,C1,D2
A4,B3,C2,D1
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by Ramesh2007 » Sun May 27, 2007 9:20 pm
Here is the matrix. First column is the letter. Second column is the matching envelopes. columns 3,4 and 5 represent the mismatching envelopes. There will be 12.
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by f2001290 » Mon May 28, 2007 12:27 am
Only one letter is in the right envelope - 4
Only two letters are in the right envelope: 4c2 = 6
Only three letters are in the right envelope: 4c3 = 4
All the envelopes are in the right envelopes: 1

4 letters can be arranged in 4 envelopes in 4! = 24 ways

Our solution = 24-(4+6+4+1) = 9 ways

Jay, I haven't yet looked at the solution sent by you. Hope I am right :wink:

But I feel sth is wrong with my 1st statement.
Last edited by f2001290 on Mon May 28, 2007 8:21 am, edited 1 time in total.
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by jayhawk2001 » Mon May 28, 2007 8:03 am
OA is 9.

Good call f2001290 and Prasanna
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by Ramesh2007 » Mon May 28, 2007 8:19 am
Thanks for the solution. Can somebody help me what I am missing in the matrix. What should I not include from this matrix while considering the answer.
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by gabriel » Mon May 28, 2007 9:07 am
f2001290 wrote:Only one letter is in the right envelope - 4
Only two letters are in the right envelope: 4c2 = 6
Only three letters are in the right envelope: 4c3 = 4
All the envelopes are in the right envelopes: 1

4 letters can be arranged in 4 envelopes in 4! = 24 ways

Our solution = 24-(4+6+4+1) = 9 ways

Jay, I haven't yet looked at the solution sent by you. Hope I am right :wink:

But I feel sth is wrong with my 1st statement.
the darkened statement shuld not be included in the calculation ... bcoz if three are in the right envelopes then the 4th will also be in the right envelope .. take a look at the solution again ..
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Let's Revisit this one, shall we?

by beeparoo » Thu May 29, 2008 12:08 pm
Let's REVISIT this one, shall we?

Posts died out on this a while ago, but the question is a good "toughie" and warrants additional minds on it.

Gabriel, guru of GMAT, exposed a flaw in the calculations by f2001290, who was luckily correct.

Prasanna's sol'n, although also correct, is too long winded to perform under tight time constraints.

Ramesh's matrix is just wrong because he should add about 6 more columns to the right in order for it to be right. He says he has 12 unique arrangements, when in fact, he has 3. The 12 he is counting is merely a sample of a letter in an envelope; these are not whole arrangements.

Can we start a fresh discussion on this one?
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by gmatinjuly » Thu May 29, 2008 12:56 pm
The answer should be 9 here is explaintation
L1 L2 L3 L4
E1 E2 E3 E4


Pick 1st letter : It can go wrong in 3 ways
Which ever slot first goes in say thats 2nd letter it can go wrong again in 3 ways

Which ever slot 2nd letter goes in .....say thats 3rd letter, now this can go wrong only in 1 way.


Example :
A. Say first letter picked is L2
L2 can fo in E1 E3 AND E4

say it goes in E3
L2
E1 E2 E3 E4


B.
So next letter is L3 , which can go in E1 E2 or E4 in 3 ways
say it goes in E2
L3 L2
E1 E2 E3 E4


C.
So now there is only one way E1 and E4 can be put wrongly.


so 3 * 3 * 1 = 9 ways all can be wrong
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by aatech » Thu May 29, 2008 12:58 pm
This is what I am thinking

First letter can go in wrong envelope in 3 ways
Second letter can go in wrong envelope in 3 ways
Third letter can go in wrong envelope in 3 ways

If 3 letters are in wrong envelope, the fourth one will be in wrong envelope so need to calculate

Total ways = 3*3 = 9
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by beeparoo » Fri May 30, 2008 1:11 pm
OK - so it seems that this question doesn't necessarily warrant any set up of permutation or combination equations.

Rather, it is more like, think-it-out-logically.

Thanks, guys.

Sandra
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