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?

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- Nov 6th 2009, 03:07 AMdukes123letter 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 AMPlato
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 AMHallsofIvy
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 AMdukes123
hey plato your answer is correct

plz explain how u got D(4)=9 its little confusing for me :( - Nov 6th 2009, 09:36 AMPlato
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 PMdukes123
thks a lot !! got to learn somethin new today :)

- Nov 7th 2009, 03:32 AMHallsofIvy
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.