Suppose that there are $n$ boxes labelled $1, 2, . . . , n$ and there are $n$ balls also labelled similarly. The balls are thrown into boxes completely randomly so that each box receives one ball. (a) How many possible arrangements of balls in boxes is possible? (b) Find the probability that the ball labelled $1$ goes into the box labelled $1$. (c) Find the probability that at least one ball is in the box with the same label.
Here is a solution.. (a) There are n balls that can be placed in $n$ boxes, so there are $n$ choices for the first ball, $n-1$ choices for the second ball, $n-2$ choices for the third ball, and so on. Therefore, the total number of possible arrangements of balls in boxes is $n!$, which is the product of all positive integers up to $n$.
(b) There are $n$ boxes, and each ball has an equal probability of landing in any of them. Therefore, the probability that the ball labelled $1$ goes into the box labelled $1$ is $\frac 1n$, since there is only one box labelled $1$ out of $n$ boxes, and each box has an equal probability of receiving the ball.
Please check my solution for $a$ and $b$ and please provide a possible solution or hint for finding $c$.
The first two parts are correctly done. For part $c$, we are just finding the number of possible derangements for $n$. You can find more about derangements of n here: https://brilliant.org/wiki/derangements/
Now for the problem we see that number of cases in which any ball does not go to it's labelled box is, $$D_n = n!(1-\frac{1}{1!}+\frac {1}{2!}-\frac{1}{3!} ... \frac{1}{n!})$$ P($n_{th}$ ball does not go into $n_{th}$ box) = $$1-\dfrac{1}{1!}+\dfrac {1}{2!}-\dfrac{1}{3!} ... \dfrac{1}{n!}$$ P(at least one $n_{th}$ ball goes into $n_{th}$ box) = $$1-(1-\dfrac{1}{1!}+\dfrac {1}{2!}-\dfrac{1}{3!} ... \dfrac{1}{n!})$$ $$=\dfrac{1}{1!}-\dfrac {1}{2!}+\dfrac{1}{3!} ... \dfrac{1}{n!}$$