<s>[INST] What is the expected number of tyres that are installed in their original positions? After summer, the winter tyres of a car (with four wheels) are to be put back. However, the owner has forgotten which tyre goes to which wheel, and the tyres are
installed `randomly', each of the $4! = 24$ permutations being equally likely.

What is the expected number of tyres that are installed in their
  original positions?

Expected no. of tyres installed in original position = $P(1\ \text{tyre}) + 2P(2\ \text{tyre}) + 3P(3\ \text{tyre}) + 4P(4\ \text{tyre})$
I'm stuck after that. When counting $P(1\ \text{tyre})$ am I allowed to take the combination that all 4 tyres are in their right position ? Because that also includes 1 tyre in its original position...
If yes then $P(1\ \text{tyre}) = 13/24$
Or must I find the probability that exactly 1 tyre is in the original position ? If so, then what is the probability that 3 tyres are in their original position ? Because having 3 tyres in the original position would mean the 4th tyre must be in the original position ?
 [/INST] $0$ tires in their correct positions (derangements):  The number of ways $k$ of the $4$ tires are in their correct positions is $\binom{4}{k}$.  The remaining $4 - k$ tires can be arranged in $(4 - k)!$ orders.  Hence, by the Inclusion-Exclusion Principle, the number of arrangements in which no tires are in their correct positions is 
$$4! - \binom{4}{1}3! + \binom{4}{2}2! - \binom{4}{3}1! + \binom{4}{4}0! = 9$$
exactly $1$ tire is in its correct position:  By the generalized Inclusion-Exclusion Principle, the number of arrangements in which exactly one tire is in its correct position is 
$$\binom{1}{1}\binom{4}{1}3! - \binom{2}{1}\binom{4}{2}2! + \binom{3}{1}\binom{4}{3}1! - \binom{4}{1}\binom{4}{4}0! = 8$$
where the factor $\binom{m}{1}$ counts the number of ways we can designate $1$ of the $m$ tires that are in their correct positions as being the one in its correct position.
exactly $2$ tires are in their correct positions:  By the generalized Inclusion-Exclusion Principle, the number of arrangements in which exactly two tires are in their correct positions is 
$$\binom{2}{2}\binom{4}{2}2! - \binom{3}{2}\binom{4}{3}1! + \binom{4}{2}\binom{4}{4}0! = 6$$
where the factor $\binom{m}{2}$ denotes the number of ways we can designate $2$ of the $m$ tires that are in their correct positions as being the ones that are in their correct positions.
exactly $3$ tires are in their correct positions: If three tires are in their correct positions, so is the fourth.  Hence, there are no arrangements with exactly three tires in their correct positions.
exactly $4$ tires in their correct positions:  There is only one arrangement in which all four tires are in their correct positions.  
Hence, the expected number of tires in their correct positions is 
$$\frac{9 \cdot 0 + 8 \cdot 1 + 6 \cdot 2 + 0 \cdot 3 + 1 \cdot 4}{4!} = 1$$
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