computing probability that at least one letter matches with envelope 
Four letters to different insurers are prepared along with
accompanying envelopes. The letters are put into the envelopes
randomly.  Calculate the probability that at least one letter ends up
in its accompanying envelope.

Attempt
Since this is tedioues, we can do
$$ P(at \; least \; one ) = 1 - P( no \; match ) $$
We notice that sample space is $4!$ since for letter 1 it has 4 choices but letter 2 has 3 choices and so on. Now, we wanna count in how many ways we get no match.
Let start with first one, we only have $3$ choices for this since it can go to either 2,3,4 envelope.
Now, as for the second one, we have to possibilities. If the first letter went to the second envelope, then the second letter now will have 3 different choices, but if the first letter didnt go to second letter, then the second letter will have 2 choices. Assume the former. Then we have 3 choices for this stage.
Now, for the third one (envelope 2 is taken already and assume letter 2 went to letter 1) then it would have 1 choice only and
last one must go to envelope 3.
Thus, we have $3 \times 3 \times 1 \times 1 = 9$ choices in total
Thus,
$$ P(at \; least \; 1 \; letter) = 1 - \frac{9}{24} $$
IS this correct? I still feel as is something wrong because I assumed the letter 2 went to 1 and letter 1 went to envelope 2. Can we do that?
 A: The probability of at least one letter ending in the correct envelope is equal to 1 - p(no letter ends up in correct envelope)
the probability of no letter ending up in its correct envelope is given by derangement. The derangement of 4 comes out to be 
$$ D(4) = 4! \left( 1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!} \right) $$ 
$$ = 9$$
Therefore, the probability of at least one letter ending in the correct envelope is 
$$  P = 1- \frac{9}{24} $$
$$  = \frac{15}{24} $$
The problem with your solution is that it does not take into account the different branches of decisions that can be made while placing letters. Such a method would ultimately reduce to just counting the cases in which derangement happens.
A: Extended Comment: This is a much-studied problem. You can find more
by searching on 'derangements' and 'hatcheck problem'.
As in SarthakNigam's answer (+1), the probability of at least
one correctly placed letter out of $n = 4$ is 15/24. An inclusion-exclusion
argument shows that a generalized alternating series works for any $n > 1.$
If $X$ is the number of correctly placed letters, then a simple
argument with indicator functions shows that $E(X) = 1,$ for all $n.$
By induction with indicator functions one can show that $Var(X) = 1,$
for $n > 1.$  
If there are more than about 10 letters, then the
distribution of $X$ is very nearly $\mathsf{Pois}(1),$ but
$P(X = n-1) = P(X > n) = 0.$
Simulation (in R) of 10 million random permutations of four letters gives
the following result (in which we get about three places of accuracy):
set.seed(419); m = 10^7;  n = 4
x = replicate(m,  sum(sample(4) == 1:4) )
table(x)/m
x
        0         1         2         4 
0.3749624 0.3331619 0.2500974 0.0417783 
mean(x > 0); mean(x);  sd(x)
## 0.6250376   # aprx P(X > 0) = 15/24 = 0.625
## 1.00047     # aprx E(X) = 1
## 1.000532    # aprx SD(X) = 1

