A probability problem - finding where I've gone wrong 
There are 7 people in an elevator and 5 floors. What is the probability of at least one person exiting each floor?

I've split this into two cases and my solution was:
$$\frac{{{7}\choose3 }5*4! + {{7}\choose 2}5{{5}\choose 2}4*3! }{5^{7}}$$
Which gives a result $0.37632$. The solution I'm provided with is $0.215$.
I assume I've gone wrong with:
$${{7}\choose 2}5{{5}\choose 2}4*3!$$
But I can't see why exactly.
I would appreciate if someone could point me to the mistake here. Thank you!
 A: In your second term, you are overcounting by a factor of two. The $\binom{7}2$ chooses one of the pairs, and $\binom{5}2$ chooses the other pair. You then assign the first pair's floor in $5$ ways, and the seccond pair's floor in $4$ ways. But it does not matter which pair was chosen first or second, so you need to divide by $2$.
Another way to see it; choose the two floors that have two people get off in $\binom{5}2=\frac{5\cdot 4}2$ ways. Then choose the people who get off on the higher floor in $\binom{7}2$ ways, and choose the people who get off on the lower floor in $\binom{5}2$ ways. Your mistake was having $5\cdot 4$ instead of $\binom{5}2=\frac{5\cdot 4}2$.
A: It is useful to check with smaller numbers: assume $5$ people ($A,B,C,D,E$) get off on $3$ floors.
The two cases are: 1) $3,1,1$ and 2) $2,2,1$.
The first case: select three people in ${5\choose 3}$ ways and distribute three units (triple, single, single) in $3!$ waysL
$${5\choose 3}\cdot 3!$$
The second case: select two people in ${5\choose 2}$ ways, select next two people in ${3\choose 2}$ ways, however, here it is double counted:
$$\begin{array}{c|c|c}
\color{red}{(AB)(CD)E} & (BC)(AD)E & \color{red}{(CD)(AB)E} & \color{green}{(DE)(AB)C}\\
\color{blue}{(AB)(CE)D} & (BC)(AE)D & (CD)(AE)B & (DE)(AC)B\\
\color{green}{(AB)(DE)C} & (BC)(DE)A & (CD)(BE)A & (DE)(BC)A\\
(AC)(BD)E & (BD)(AC)E & (CE)(BD)A & \\
(AC)(BE)D & (BD)(AE)C & \color{blue}{(CE)(AB)D} & \\
(AC)(DE)B & (BD)(CE)A & (CE)(AD)B & \\
(AD)(BC)E & (BE)(AC)D & & \\
(AD)(BE)C & (BE)(AD)C & & \\
(AD)(CE)B & (BE)(CD)A & & \\
(AE)(BC)D & & & \\
(AE)(BD)C & & & \\
(AE)(CD)B & & & \\
\end{array}$$ 
Note: Only $3$ pairs are highlighted, whereas there are $15$ pairs in total. 
Hence, in the second case:
$$\frac{{5\choose 2}{3\choose 2}}{2}\cdot 3!.$$
