Question 696876
<pre>
When you see the words "at least one", we know to first find the probability of
the complement event and then subtract from 1.

The complement event is that no letters go in the right envelope.

The denominator of the probability is obviously 5!.

The numerator of the probability is what is difficult.

Let a<sub>n</sub> be the number of ways n letters can all put in n wrong envelopes.

We want to find a<sub>5</sub>  

Suppose the 5 letters are numbered 1,2,3,4,5, and the envelopes are also
numbered 1,2,3,4,5. We have to find the number of ways in which no letter goes
in the envelope having the same number as its number. There are 4 ways for the
first letter to go in an envelope other than the 1st one.  Let us assume, for
definiteness, that the letter 1 gets put in envelope 3.  There are now two
possibilities:

case 1. letter 3 goes in envelope 1,

That leaves letters 2,4,5 to be put wrongly in envelopes 2,4,5.  That's
a<sub>3</sub> possibilities.

case 2. letter 3 does not go in envelope 1.

That leaves letters 2,3,4,5 to put wrongly in envelopes 1,2,4,5. But since
letter 3 cannot go in envelope 1, that will be the same as if we re-labeled
letter 3 as letter X and re-labeled envelope 1 as envelope X. Then we would
have letters 2,X,4,5 to go wrongly in envelopes X,2,4,5.  That's a<sub>4</sub>
possibilities.
   
So the number of ways letter 1 could have gone in envelope 3 is 
a<sub>3</sub> + a<sub>4</sub> 

But we could have put letter 1 in any of the 4 letters {2,3,4,5}.  Therefore we
must multiply that by 4, and so we get

a<sub>5</sub> = 4(a<sub>3</sub> + a<sub>4</sub>)

By the same token

a<sub>4</sub> = 3(a<sub>2</sub> + a<sub>3</sub>) 

It's obvious that a<sub>2</sub> = 1 because if there are 2 letters and 2
envelopes, then there is exactly 1 way to put letter 1 in envelope 2 and letter
2 in envelope 1.

It's also easy to discover that a<sub>3</sub> = 2 because envelopes 1,2,3 could
only either go respectively in envelopes 2,3,1 or respectively in envelopes
3,1,2.

Therefore, 

a<sub>4</sub> = 3(a<sub>2</sub> + a<sub>3</sub>) = 3(1+2) = 3(3) = 9

and

a<sub>5</sub> = 4(a<sub>3</sub> + a<sub>4</sub>) = 4(2+9) = 4(11) = 44

So the probability that all the letters will go wrong is 44 ways out of all 5!
ways they could have been placed. That probability is {{{44/5!}}} = {{{44/120}}} = {{{11/30}}}.

Therefore the probability that at least one letter goes in the correct envelope
is 1 - {{{11/30}}} = {{{19/30}}}

Edwin</pre>