Originally Posted by

**Jhevon** I take it [7] = {1,2,3,4,5,6,7} and [9] = {1,2,3,4,5,6,7,8,9} ...(?)

**Yes this is correct**

why are you adding? you should be multiplying...

**Consider [3]$\displaystyle \longrightarrow$[4] with f(2)=3**

We then have {1,3} $\displaystyle \longrightarrow$ {1,2,3,4}. This gives $\displaystyle 4^2$ but then we have the additional f(2)=3. This would be added on wouldn't it?

I am not familiar with the notation you are using. $\displaystyle (9)_6 = _9C_6$?

**No this stands for $\displaystyle _9P_6$**

Anyway, I would disagree. We have 8 choices for what to map 3 to. Having made that choice, we again have 8 choices to map the second element to, then 7, then 6, then 5, etc. The total number of functions with this criteria, therefore, is $\displaystyle 8^2 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3$

**Then I would need to multiply by the 8 instead of adding it and would agree except for the final factor of 3**

I agree

what is your definition of $\displaystyle f^{-1}$ here? Because, unless we restrict the domain somehow, the inverse will technically never be defined.