<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] *

*Probability that exactly 2 repairers are called.
There are two ways how two repairers are called:


*

*Both gets call for two broken sets each
First repairer $A$ can be anyone. Hence, will have probability $\frac{4}{4}=1$. The same repairer will be called for any one of 2nd, 3rd or 4th broken TV set. [Hence the total possible sequences can be 3 $(AABB,ABAB,ABBA)$.] At that time $A$ can be selected with probability $\frac{1}{4}$, that is, he should be same as first one. The 2nd repairer can be any one apart from $A$, Thus, the first call to 2nd repairer $B$ can happen with probability $\frac{3}{4}$. The 2nd call to $B$ can happen with probability $\frac{1}{4}$. Final probability $\frac{4}{4}\times\frac{3}{4}\times\frac{1}{4}\times\frac{1}{4}\times3$

*One gets call for three broken TV sets, while the other one gets call for 
remaining one broken set, then there are four ways:
$(AAAB,AABA,ABAA,BAAA)$.
Each one will have same probability as first bullet point. Hence final probability: $\frac{4}{4}\times\frac{3}{4}\times\frac{1}{4}\times\frac{1}{4}\times 4$. 
Final probability of 1st and 2nd bullet point combined: $\frac{4}{4}\times\frac{3}{4}\times\frac{1}{4}\times\frac{1}{4}\times 7 =\frac{21}{64}$


*Probability that exactly 3 repairers are called.
Let the three repairmen called be $A,B,C$.
There can be six sequences in which call are made for 1st, 3nd, 3rd and 4th broken TV sets: $(ABCA,ABAC,AABC,ABBC,ABCB,ABCC)$. First repairman can be called with probability $\frac{4}{4}$. Second new repairman can be called with probability $\frac{3}{4}$. Third new repairman can be called with probability $\frac{2}{4}$. Any repairman can be repeated with probability $\frac{1}{4}$. Final probability $=\frac{4}{4}\times\frac{3}{4}\times\frac{2}{4}\times\frac{1}{4}\times 6=\frac{36}{64}$

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