How many injective functions $f$ are there from set $\{1, 2, . . . , n\}$ to the set $\{1, 2, . . . , n\}$ so that $f(x) = x$ for some $1 ≤ x ≤ n$?

Question

Using Inclusion/Exclusion and to solve this question, I am confused from the word "some" in the wording.

So far, I have determined the inclusion part to be the total number of injective functions with no constraints. This is $\boldsymbol {n!}$. Is the exclusion for the constraint just $\boldsymbol{-1}$ so that the total number of functions is $\boldsymbol{n! - 1}$?

Any guidance would be great.

Answer 1

The some means there is at least one $x_i$ such that $f(x_i)=x_i.$

If you want to do inclusion/exclusion you could say that there are $(n-1)!$ functions where $f(1)=1$ because you can order the rest of the numbers any way you want. You have $n$ choices of the fixed point, so this would give $n((n-1)!)=n!$ functions, but you have counted the ones with two fixed points twice so have to subtract them once. Then the ones with three fixed points have been counted three times and removed three times, so have to be added once.

The approach in N. F. Taussig's comment is easier.