Find the number of $4$-letter words using $A,B,C,D,E$ with the following conditions I`m trying to find how much words in length $4$ I can create from $A,B,C,D,E$; so
the conditions are:


*

*The number of words that have to be at least with the letter $A$.

*The number of words that have to be at least with the letter $A$ and
other letters aren't allowed to be the same, for example $ABCD$ is possible, but not $ABBB$.


Until now what I think is $5^3 \cdot 4$, can someone give a suggestion?
 A: First, figure out how many words you could create if you did not need at least 1 A, i.e. the questions would be:


*

*How many possible words four letters long can you create with the letters {A, B, C, D, E}?

*How many possible words four letters long can you create with the same letters, but using each letter at most once?
Once you've done this, you want to figure out how many of those have at least 1 A, but, as Gerry has pointed out, this can easily be done by figuring out how many possible 4-letter words of the appropriate type you could create by using only {B, C, D, E}, because if you take all possible words created using A-E, and then remove all the words that use only B-E, you're left with all the words that have at least 1 A (otherwise they'd be removed).
A: I think that the most elegant way to describe the solution is to consider the real number of letters that can occupy a certain "position" on the spots available; since you told that the words contains at least the letter A, I'd like to count all the possible ways that this letter could appear in four spaces, supposing that it could repeat:
$$
N_{A}=\underbrace{P_4^{(1,3)}}_{A - - -}+\underbrace{P_4^{(2,2)}}_{AA - -}+\underbrace{P_4^{(3,1)}}_{AAA-}+\underbrace{P_4^{(4)}}_{AAAA}
$$ 
Now inside each possibility shown the other letters, except A can appear without repetition and taking into account the order, so it just needs a "filling" with all the remaining letters:
$$
N_{\mathrm{tot}}=\underbrace{P_4^{(1,3)}\cdot D_{4,3}}_{A - - -}+\underbrace{P_4^{(2,2)}\cdot D_{4,2}}_{AA - -}+\underbrace{P_4^{(3,1)}\cdot D_{4,1}}_{AAA-}+\underbrace{P_4^{(4)}}_{AAAA}
$$
Now the number of words needed (if I understood the request correctly) satisfy the two initial conditions, giving the number:
$$
N_{\mathrm{tot}}=\frac{4!}{3!}\cdot (4\cdot 3\cdot 2)+\frac{4!}{2!\cdot 2!}\cdot (4\cdot 3)+\frac{4!}{3!}\cdot (4)+1=16\cdot 6+4\cdot 18+16+1=185
$$
I notice that this particular answer was constructed in a way to eliminate by progression the possibility of the "filling" by other remaining letters, excluding the first request, ergo the letter A in the first place.

For the second condition, the number of A's can not be greater than the first, because it's not taken into account the repetition; so considering the order of the letters the shorter method takes all the possible letters inserted in any order:
$$
N_{\mathrm{tot}}=\underbrace{5}_{\ell---}\cdot \underbrace{4}_{-\ell--}\cdot \underbrace{3}_{--\ell-}\cdot \underbrace{2}_{---\ell}=120
$$
In this case is assured that A is picked only one time because it's position is not fixed, but inside all of $120$ possibilities there are exactly:
$$
\underbrace{P_4^{(1,3)}\cdot D_{4,3}}_{A - - -}
$$
Those words only contain A, so the total number needed is:
$$
N_{\mathrm{A}}=\underbrace{P_4^{(1,3)}\cdot D_{4,3}}_{A - - -}=\frac{4!}{3!}\cdot (4\cdot 3\cdot 2)=16\cdot 6=96
$$
As advised in this more restricted case.

If in the first place the letters $B-E$ can be repeated, then it will be added the number of cases that can exist using more than one letter per position available, resulting in one Repeated Distribution ($D'_{n,k}$):
$$
N_{\mathrm{tot}}=\underbrace{P_4^{(1,3)}\cdot D'_{4,3}}_{A - - -}+\underbrace{P_4^{(2,2)}\cdot D'_{4,2}}_{AA - -}+\underbrace{P_4^{(3,1)}\cdot D'_{4,1}}_{AAA-}+\underbrace{P_4^{(4)}}_{AAAA}
$$
Transforming the number obtained by:
$$
N_{\mathrm{tot}}=\frac{4!}{3!}\cdot (4^3)+\frac{4!}{2!\cdot 2!}\cdot (4^2)+\frac{4!}{3!}\cdot (4)+1= 4^4 + 4^2\cdot 6 + 4^2+1=369
$$
Now that the first case is extended, it has been shown that one simple equation for $N_{\mathrm{tot}}$ can be stretched with the other conditions posed by the problem.
