Fixed solution to 1.11

Surely if two sets are finite and have the same number of elements, there exists an injection between them?

Author: wendawu158

src/chapters/1/sections/counting/problems/11.tex

@@ -2,10 +2,10 @@
 \item Each of the $n$ inputs has $m$ choices for an output, resulting in 
 $$m^{n}$$ possible functions.
 
-\item If $n \geq m$, at least two inputs will be mapped to the same output, 
+\item If $n > m$, at least two inputs will be mapped to the same output, 
 so no one-to-one function is possible. 
 
-  If $n < m$, the first input has $m$ choices, the second input has $m - 1$ 
+  If $n \leq m$, the first input has $m$ choices, the second input has $m - 1$ 
   choices, and so on. The total number of one-to-one functions then is 
   $$m(m-1)(m-2)\dots(m-n+1) = \frac{m!}{(m-n)!} $$
-\end{enumerate}
+\end{enumerate}