The answer should be 6!.

The first letter can be put into any one of 6 envelopes, the second letter can be put into any one of 5, the 3rd into any one of 4 and so on.

Can someone help me clarifying my answer.

Actually when I'm trying to visualize a smaller set, say of 3, then, I feel my answer is wrong.

I want to know if my approach is correct.

Thanks for this concept. I just did google search and found a whole lot of material. That'll be a real help. Thank you!

Solution:

Why are you asking this question? First of all, IMO, although this problem looks basic, in reality this problem is WAY WAY WAY too difficult for the GMAT and will stump even graduate math students (it took me about 30 minutes). IMO, asking problems outside the bounds of GMAT knowleged is counter productive because one of the "tricks" to solving GMAT problems is the concrete knowledge that every problem, no matter how complex at first sight, is solvable using basic maths. Hence, you build confidence by knowing that there must be a trick to solving this in a short amount of time without esoteric knowledge.

If you are going to post a question that you knowingly know is outside the boundaries, you should let everyone know so that they don't waste time looking for that illusive "insight" that is not there to find...

Here is my solution. If you can find a simpler solution that can be executed in 2 minutes without special math knowledge, I take everything I said above back and will kneel "i'm not worthy" to you 100 times.

We are looking for the solution of G(6) where G(n) = number of ways n items can be permuted such that they none of them match their original position. This is the same as having a deck of n cards numbered from 1 to n, then counting the ways you can shuffle the deck without any cards matching its position.

Consider this: it is obvious that when n = 1, G(n) must be ZERO, and when n = 2, G(n) = 1 (i.e., {2,1}), and when n = 3, G(n) = 2 (i.e., {2,3,1},{3,1,2}). By drawing out a probability tree, you will discover that G(4) = 9, and with a LOT of work and patience, G(5) = 44.

If you carefully watch how the tree is growing, you may notice that the results of G(n) refer back to the results of G(n-1) and G(n-2). With a little bit of work, we can see that a general recursive formula for G(n) is:

G(n) = (n-1)*(G(n-1)+G(n-2)) where n > 1 and G(1) = 0 and G(2) = 1.

I was not able to find a closed form solution for any n.

If you build a table, then G(6) comes out to 5*(44 + 9) = 265. Since there are 6! or 720 ways to stuff the envelopes, IMO, the probabiltiy of getting them all wrong is 265/720 = 53/144.
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

there are 720 possible arrangements.

1/720 where they are all correct.
0/720 where 5/6 are correct.
15/720 where 4/6 are correct

***there are 6c3 or a total of twenty ways for three to be correct, and in each of those ways, the wrong configuration can happen in two different ways.*** therefore,

40/720 where 3/6 are correct.

***there are 6c2 or a total of fifteen ways for two to be correct, and in each of those ways, the wrong configuration can happen in nine different ways.*** therefore,

105/720 where 2/6 correct.

***there are 6c1 or a total of six ways for one to be correct, and in each of those ways, the wrong configuration can happen in 40 different ways.*** therefore,

240/720 where 1/6 correct.

401 total where 1/6 correct, 319 where none are correct.

From reading this thread, I bet I'm wrong, and I think I erred in the last part where I figured out how many ways the wrong configuration can happen with one right. The remainder of my work took very little time, but that last step was a bitch.