In how many ways can we distribute $20$ cards among $20$ persons numbered from $1$ to $20$ such that no one get his own number?

Question

In my opinion inclusion-exclusion formula could be used here,but I think there must be a better way so we can reach the final answer(in fewer terms). Do you have any idea?

Answer 1

Its problem of typical dearrangement whose total ways are given by $$n![1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}...+\frac{(-1)^n}{n!}]$$

Answer 2

You probably would have a really difficult time calculating this with the inclusion-exclusion formula.

The answer is $895014631192902121$, as proved by the reucrrence relation for number of derangements for $n$, or the general formula, which is $!n=\lfloor \frac{n!}{e}+\frac{1}{2} \rfloor$

If you are curious what the recurrence relation is, see here and here.