# Let F be a set of $1$-to-$1$ functions from the set $\{1,2,....,n\}$ to the set $\{1,2,....m\}$ where $m \geq n \geq 1$?

How many functions are members of $F$?

$$\dfrac{m!}{(m - n)!}$$

But they said i was wrong and the answer is $mn$

Where am i wrong?

• Your answer is indeed correct. "They" probably missed the fact that F contains only one-one functions. If any functions were allowed, then "they" are right. Jan 14, 2014 at 4:58

We have $m$ choices for what $1$ is sent to. For each of these, we have $m-1$ choices for what $2$ is sent to. For each way of doing these two things, there are $m-2$ choices for what $3$ is sent to, and so on for a total of $$m(m-1)(m-2)\cdots(m-n+1).$$ This can also be written as $\frac{m!}{(m-n)!}$.

Your calculation is correct. The answer $mn$ is not, unless the problem was incorrectly described.

Who is this "they"? You first pick the range of your function (can be done in $\binom{m}{n}$ ways, then any permutation thereof gives you a function (and a different one), giving your formula, and not "theirs".

• It's always "they" who create problems by dictating whether we are right or wrong regardless of our method. Jan 14, 2014 at 4:59
• appreciated for the support. Jan 14, 2014 at 5:04