7 distinguishable balls in 4 distinguishable boxes, probability that no box is left empty. I found a solution for this that uses multinomial however I tried to solve it in a different way and I don't know where I am going wrong.
So I thought how about we fill each box with one ball first, and then put the remaining 3 in all possible boxes. 
This was my reasoning:
First let's choose 4 balls from the 7 and order them, that's 7C4*4! 
Now the number of ways to place the remaining 3 balls is: 4^3 
The cardinality of our sample space is 4^7 
So the probability should be: (7C4*4!*4^3)/4^7 
However this yields a number greater than 1. 
Could you point where I am going wrong?
 A: Use inclusion/exclusion principle.

First, count the number of ways to distribute $7$ balls into $4$ boxes so that no box is empty:


*

*Include the number of ways to distribute $7$ balls into at most $\color\red4$ boxes, which is $\binom{4}{\color\red4}\cdot\color\red4^7$

*Exclude the number of ways to distribute $7$ balls into at most $\color\red3$ boxes, which is $\binom{4}{\color\red3}\cdot\color\red3^7$

*Include the number of ways to distribute $7$ balls into at most $\color\red2$ boxes, which is $\binom{4}{\color\red2}\cdot\color\red2^7$

*Exclude the number of ways to distribute $7$ balls into at most $\color\red1$ boxes, which is $\binom{4}{\color\red1}\cdot\color\red1^7$


Then, divide this result by the total number of ways to distribute $7$ balls into $4$ boxes:
$$\frac{\binom{4}{4}\cdot4^7-\binom{4}{3}\cdot3^7+\binom{4}{2}\cdot2^7-\binom{4}{1}\cdot1^7}{4^7}$$

Please note that you get the same probability even if the balls and/or the boxes are indistinguishable from each other.
A: This answer uses Inclusion-Exclusion just like barak manos' answer, but as the description in that answer seems a bit confusing, I am posting this answer.

The number of ways for one particular box to be empty (other boxes may be empty)
$$
\overbrace{\ \ \ \binom{4}{1}\ \ \ }^{\substack{\text{number of ways to choose}\\\text{the particular box}}}\,\,\overbrace{\ \ \ \ \ \vphantom{\binom{4}{3}}3^7\ \ \ \ \ }^{\substack{\text{number of ways to put the $7$ balls}\\\text{into the remaining boxes}}}
$$
The number of ways for two particular boxes to be empty (other boxes may be empty)
$$
\binom{4}{2}2^7
$$
The number of ways for three particular boxes to be empty
$$
\binom{4}{3}1^7
$$
Inclusion-Exclusion says that the number of ways to have at least one box empty is
$$
\binom{4}{1}3^7-\binom{4}{2}2^7+\binom{4}{3}1^7
$$
Since there are $4^7$ ways to put the balls in the boxes, there are
$$
4^7-\binom{4}{1}3^7+\binom{4}{2}2^7-\binom{4}{3}1^7
$$
ways to have no boxes empty. The probability of this is
$$
\frac{4^7-\binom{4}{1}3^7+\binom{4}{2}2^7-\binom{4}{3}1^7}{4^7}\doteq51.27\%
$$
A: Different approach, different solution:
Count arrangements (for 'no box empty' I don't understand why 'distinguishable' matters)
A) All arrangements: place 3 bars between 7 balls: $10 \choose 3$.
B) Non-empty arrangements: Then remove 4 balls, one for each box, and place 3 bars between 3 items: $5 \choose 3$
Probability: B) / A) = $\frac{1}{12}$
Then, when I look at the reputation of the other authors, I wonder where I went wrong.
(Or are not all arrangements equiprobable?)
