a question on probability where in a pack of n cards, numbered 1, 2, ..., n, is shuffled and laid out in a row. A pack of n cards, numbered 1, 2, ..., n, is shuffled and laid out in a row. The result of the shuffle is that each card is equally likely to be in any position in the row.
Hence show that the probability that at least one card is in the same position as the number it bears is

I tried to do it but couldn't get anywhere close to the solution.
 A: Let $a_n$ be the event that the $n$-th card is in its position. We wish to find the probability that one card, two cards,...or $n$ cards are in their position. Notationally that is,
$$\mathbb{P}(a_1 \bigcup a_2 \bigcup ...\bigcup a_n)=\mathbb{P}(\bigcup_{k=1}^{n} a_k)$$

By the principle of inclusion exclusion this is the same as adding up the probabilities of each event occurring , then subtracting the probabilities of the double intersections occurring, then adding the probabilities of the triple intersections occurring, Etc. 
Because there are $n$ cards we repeat this until we get to the "$n$"-th order intersections.
We will calculate each of the probabilities necessary to utilize inclusion exclusion.

$$\mathbb{P}(a_1)$$
The probability a specific card is in the same place is $\frac{1}{n}$ because each individual card is as likely to be in any of the $n$ positions. 
Note: This means that,
$$\mathbb{P}(a_1)+\mathbb{P}(a_2)+...\mathbb{P}(a_n)=n\frac{1}{n}=1$$

$$\mathbb{P}(a_1 \bigcap a_2)$$
The probability that one card and another card is in the same position is $\frac{(n-2)!}{n!}=\frac{1}{n(n-1)}$ because two cards must be in their positions (there is only one way to get each in their position) and the rest $n-2$ can be in any order. There are $n!$ ways to order $n$ cards and $(n-2)!$ ways to order $n-2$ cards. Hence by the multiplication principle the number of ways $a_1$ and $a_2$ can happen is $(1)(1)(n-2)!$ out of a total of $n!$ possible ordering a for the cards.
Note:
There are ${n \choose 2}=\frac{n(n-1)}{2}$ ways to get "double intersections".

$$\mathbb{P}(a_1 \bigcap a_2 ....\bigcap a_n)$$
This probability, the probability that all cards are in their same place is $\frac{1}{n!}$ because there is only $1$ way to get that out of $n!$ possibilities. There are ${n \choose n}=1$ "$n$" intersections.

Combining the results by inclusion-exclusion gives what we want,
$$\frac{1}{n}n-\frac{(n-2)!}{n!} {n \choose 2}+\frac{(n-3)!}{n!} {n \choose 3} \cdots +(-1)^{n+1}\frac{(n-n)!}{n!} {n \choose n}$$
