# Montmorts Problem on random arrangements

So given A = "neither student 1 nor student 2 gets put in their own desk"

X = indicator function for A

Y = the number of students do not get put back on their own desk

part 1 lets assume n = 10 so there were 10 students and 10 chairs Sample space would be = {(Sn,Cn)} where n is ranged from 1 t0 10

i dont know how to find the range and i have problems with part 2

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To show formally that $X$ and $Y$ are not independent, consider the event $(X=1)\cap (Y=0)$. This event has probability $0$, since if $Y=0$, then everybody gets put back at her/his own
But neither $\Pr(X=1)$ nor $\Pr(Y=0)$ is $0$. Thus $$\Pr((X=1)\cap (Y=0))\ne \Pr(X=1)\Pr(Y=0).$$
As to the first question, there is some unexplained notation that I would not hazard a guess about. But on the assumption that $f_X$, as usual, represents the distribution function of $X$, we can proceed as follows. The random variable $X$ only takes on the values $1$ and $0$. We find $f_X(1)$, the probability that $X=1$. If $X=1$, then either (i) Student 1 got moved to 2's desk or (ii) to someone else's.
The probability that Student 1 got moved to 2's desk is $\frac{1}{n}$. If that happened, then for sure 2 will not end up at her own desk.
The probability Student 1 got "moved" to a desk other than her own or 2's is $\frac{n-2}{n}$, Given that happened, the probability that 2 doesn't get moved to her own desk is $\frac{n-2}{n-1}$. Thus $$\Pr(X=1)=\frac{1}{n}+\frac{n-2}{n}\cdot \frac{n-2}{n-1}.$$