Number of ways so that two janitors do not have work to do A university has recently hired 5 janitors after noticing that toilets were consistently being clogged by paper. There are 23 toilets that are currently overflowing. How many ways can we assign the janitors to the toilets so that exactly two janitors have no work to do?  
This is what I thought:
Let $x_i$ be the number of toilets that the $i$th janitor is working on. So we count the number of positive integer solutions to
$$x_1 + x_2 + x_3 + x_4 + x_5 = 23$$
where exactly one of $x_j,x_k$ for $j\neq k$ are zero.  
So we pick two janitors in $\binom{5}2$ ways.
Then it comes down to counting number of positive integer solutions to
(WLOG)
$$x_1 + x_2 + x_3 = 23$$
which is done in $\binom{22}{2}$ ways.
So total is $\binom{5}2\binom{22}2$. However, the answer key gives a much larger solution.
Why is my way incorrect?
 A: We are now going to use the formula to determine when exactly two conditions we impose are met. Let our condition $c_j $ be defined by janitor $j$ is doing no work. So we are looking for $E_2$, this is, when exactly two of the janitors are doing no work. This is given by 
$$E_2 = S_2 - \binom{3}{1}S_3 + \binom{4}{2} S_4 - \binom{5}{3}S_5   $$
Now, who are $S_2,S_3,S_4 $ and $S_5$?


*

*$S_2 = \binom{5}{2}3^{23}$  means that we can pick 2 janitors to do no work at all, and so the 23 toilets will each be assigned to either of three janitors, hence the $3^{23}$.

*$S_3 =  \binom{5}{3}2^{23}$ means that we now pick 3 janitors to do no work, and then the 23 toilets will each be cleaned by one of the two janitors who are working, hence the $2^{23}$.

*$S_4 =  \binom{5}{4}$ means that we now pick 4 janitors to do no work, and then the 23 toilets will all be cleaned by the same janitor. Hence we would have $1^{23} = 1.$

*$S_5 =  \binom{5}{5} \cdot 0 = 0$ would mean when none of the janitors choose to work.


So now we plug in those values: 
$$E_2 = \binom{5}{2}3^{23} - \binom{3}{1} \binom{5}{3}2^{23} + \binom{4}{2}  \binom{5}{4}  $$
