# letter in envelope problem

• Nov 6th 2009, 03:07 AM
dukes123
letter in envelope problem
i not getting the correct answer for this sum : In how many ways can five letters be placed in five envelopes such that exactly one letter goes into corresponding envelope?
• Nov 6th 2009, 03:18 AM
Plato
Quote:

Originally Posted by dukes123
i not getting the correct answer for this sum : In how many ways can five letters be placed in five envelopes such that exactly one letter goes into corresponding envelope?

It is hard to know exactly what you are asking.

But this may be it.
A derangement is rearrangement of a collection in which no element remains fixed.
So you can chose one of five to be fixed and then derange the other four.
So, $\displaystyle 5\cdot D(4)=5(9)=45.$
• Nov 6th 2009, 04:37 AM
HallsofIvy
Quote:

Originally Posted by dukes123
i not getting the correct answer for this sum : In how many ways can five letters be placed in five envelopes such that exactly one letter goes into corresponding envelope?

Do you mean exactly one letter goes into each envelope?

You have 5 choices which letter to put into the first envelope, then 4 choices which to put in the second envelope, then 3, then 2, then 1. 5*4*3*2*1= 5!

This is exactly the same as the number of different ways to order 5 objects.

Plato's answer was assuming that each letter had a correct envelope and you were asking "how many ways can you put 5 letters in 5 evelopes if no letter goes into the correct envelope". I did not interpret your question that way.
• Nov 6th 2009, 08:28 AM
dukes123
plz explain how u got D(4)=9 its little confusing for me :(
• Nov 6th 2009, 09:36 AM
Plato
Quote:

Originally Posted by dukes123
plz explain how u got D(4)=9 its little confusing for me :(

Here is a very helpful website for derangements

Here is a formula: $\displaystyle D(n) = n!\sum\limits_{k = 0}^n {\frac{{( - 1)^k }}{{k!}}}$.

Here is a good approximation: $\displaystyle n \geqslant 4\, \Rightarrow \,D(n) \approx \frac{n!}{e}$.

$\displaystyle D(4)=\frac{4!}{e}\approx 8.8291$, so $\displaystyle D(4)=9$
• Nov 6th 2009, 09:59 PM
dukes123
thks a lot !! got to learn somethin new today :)
• Nov 7th 2009, 03:32 AM
HallsofIvy
Oh, I see now. "exactly one letter goes into corresponding envelope" means exactly one letter goes into the correct envelope and the other four into the wrong envelopes.