The number of functions $ f : \left \{ 1, 2, . . . , 10 \right \} \rightarrow \left \{ 1, 2, . . . , 10 \right \}$ such that $f(x) \neq x$ for all $x$

Question

Question

The number of functions $ f : \left \{ 1, 2, . . . , 10 \right \} \rightarrow \left \{ 1, 2, . . . , 10 \right \}$ such that $f(x) \neq x$ for all $x$ is

Approach

Total number of function possible $=10^{10}$

But with the restriction ,$f(x) \neq x$ for all $x$,

Eg-:$f(1)\neq 1 $ each element in the domain will have 1 less available option in the range .

So total number of function possible$=9^{10}$

Am I correct?

please help.

Answer 1

For every input, you have 9 choices for output. As $f(x)=x$ is not allowed. So, total no. of ways is $9^{10}$

Answer 2

Case 1: If functions are not bijective.

It's a simple case for every $i^{th}$ element ,$1\le i\le 10$, you have 9 possibilities to choose from for the output.

Hence total number of functions=$ 9^{10}$

Case 2: If functions are bijective

On close observation you might notice that what you need is exactly the number of Derangements that could be done with $10$ elements. That means you need to find $$!10=\left\lfloor \frac {9!}{e}+\frac 12\right\rfloor =362 880 $$