<s>[INST] How to count permutations with restrictions on how items are grouped I am trying to solve the following problem:

A town contains $4$ people who repair televisions. If $4$ sets break down, what is the probability that exactly $i$ of the repairers are called? Solve the problem for $i=1,2,3,4$.

For $i=1$, there are ${}_4P_1$ ways to assign $1$ person to $4$ televisions, so the probability is $\frac{{}_4P_1}{4^4}=\frac{1}{64}$.
For $i=4$, there are ${}_4P_4$ ways to assign $4$ people to $4$ televisions, so the probability is $\frac{{}_4P_4}{4^4}=\frac{3}{32}$.
I am having trouble with $i=2,3$. How should I go about these cases?
 [/INST] Conditions:


*

*The repairmen and TVs are distinct.

*Every TV owner calls ONLY one repairman, that is, the relationship between the set of TVs, $T$, and a set that contains $i$ of the repairmen, $Ri$, is functional, $f: T -> Ri$. Note that $|T|=4$ and $|Ri|=i$

*Every one of the $i$ repairmen gets at least one call, that is the range of $f$ is $Ri$.


The size of the sample space is $|S| = 4^4 = 256$.
$i = 2$:
There are $\binom{4}{2}=6$ ways to select two repairmen. Due to conditions #2 and #3, each repairman can get 1 OR 2 OR 3 calls, for the other will get the remaining calls; therefore, there are $\binom{4}{1} + \binom{4}{2} + \binom{4}{3} = 14$ ways for each repairman to get calls. 
You can also argue as follows to obtain the same result: To satisfy conditions #1, #2 and #3, the number of calls can be divided between the two repairmen in one of the following ways: 1|3, 2|2, or 3|1; This is the same way we must partition the set of TVs. Using multinomial coefficients, these partitions can be counted in $\binom{4}{1,3}+\binom{4}{2,2}+\binom{4}{3,1}=14$ ways.  
Hence $P\{i=2\} = (6*14)/256 = 84/256$.
$i = 3$:
There are $\binom{4}{3} = 4$ ways to select the repairmen. A repairman can get 1 OR 2 calls, for the other two repairmen will get the remaining calls.  
If a repairman gets 1 (out of 4) call, the second repairman can get either 1 OR 2 (out of remaining 3) calls, and the third repairman gets the remaining calls. This can be counted in$\binom{4}{1}\binom{3}{1} + \binom{4}{1}\binom{3}{2}=24$ number of ways. 
If a repairman gets 2 (out of 4) calls, the second repairman can only get 1 (out of 2) call, and the third repairman gets the remaining call. This can be counted in $\binom{4}{2}\binom{2}{1}=12$ number of ways. 
Therefore, the total number of ways the repairmen are called is $24+12=36$.
The same result can be more succinctly obtained by multinomial coefficients: $\binom{4}{1,1,2}+\binom{4}{1,2,1}+\binom{4}{2,1,1}=36$.
Hence $P\{i=3\} = (4*36)/256 = 36/64$.
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