What would be the expected number of targets which didn't get hit by any of the shooters? 
Assume there are 5 shooters and 5 targets. Each shooter would choose a target randomly and when the signal is given all shooters will fire at the same time and hit their target. What would be the expected number of targets which didn't get hit by any of the shooters?

My attempts: Initially, there are $5^5$ different possible case. Now I assume a R.V $X$ be the number of unshot target. So $P(X=0)=5!/5^5, P(X=1)=5!/5^5,P(X=2)=C_2^5\times3!/5^5$ and so on but it seems that the probability I get is wrong but i am not sure which part goes wrong and how to get the correct one. Any solution or hints are welcome, thanks.
 A: Define random variables $X_1,X_2,\dots, X_5$ by $X_i=1$ if Target $i$ didn't get hit, and by $X_i=0$ otherwise. Then the number of targets hit is $X_1+\cdots+X_5$. By the linearity of expectation, 
$$E(Y)=E(X_1)+E(X_2)+\cdots+E(X_5).$$
To find $E(X_i)$, we only need to find the probability Target $i$ didn't get hit. This is (if the shooters choose their targets uniformly and independently) equal to $\left(\frac{4}{5}\right)^5$. This is $E(X_i)$. 
Remark: Note that we do not need to know the distribution of $Y$ to calculate the expectation. The $X_i$ are not independent, but linearity of expectation always holds.  Also, the approach works just as smoothly with $k$ targets and $n$ shooters. 
A: Your approach is the hard way, but it can be done.
Consider the case of one missed target. There are $5$ ways to choose which target is missed. There are $\binom52$ ways to choose which $2$ of the $5$ shooters hit the same target, and there are $4$ ways to choose which target it is. There are then $3!$ ways to assign targets to the remaining $3$ shooters, so there are altogether
$$5\cdot\binom52\cdot4\cdot3!=\binom525!=10\cdot5!$$
ways to get $X=1$. Thus,
$$\Bbb P(X=1)=\frac{10\cdot5!}{5^5}=\frac{2\cdot4!}{5^3}\;,$$
not $\dfrac{5!}{5^5}$.
For two missed targets there are $\binom52$ ways to choose the missed targets, but the rest of the calculation is a bit more complicated, because there are two possibilities.


*

*We can have $3$ shooters hitting one target. There are $\binom53$ ways to choose the shooters and $3$ ways to choose their target, and there are then $2$ ways to assign the other two targets to the other two shooters.  

*We can have two targets hit by two shooters each. There are $\binom32$ ways to choose which two targets will be hit twice. There are $\binom52$ ways to choose which $2$ shooters will hit the first of these, and there are then $\binom32$ ways to choose which $2$ will hit the other one.


The total count of ways to get $X=2$ is therefore
$$\binom52\left(\binom53\cdot3\cdot2+\binom32\binom52\binom32\right)=1500\;,$$
and 
$$\Bbb P(X=2)=\frac{1500}{5^5}=\frac{12}{25}\;.$$
This analysis can be continued, but you can see that it will be quite a bit of work.
The easy approach is to use linearity of expectation on $X=\sum_{k=1}^5X_k$, where $X_k$ is $1$ if target $k$ is missed and is $0$ otherwise; I see that while I was writing up the messy bit, André has already covered this.
