Maps $g$ maps $\left\{1,2,3,4,5\right\}$ onto $\left\{11,12,13,14\right\}$ and $g(1)\neq g(2)$. How many g are there. Maps $g$  maps $\left\{1,2,3,4,5\right\}$ onto $\left\{11,12,13,14\right\}$ and $g(1)\neq g(2)$. How many g are there.
My answer:
I transformed the question to a easy-understand way and find out the solution.
Consider there are five children and four seats. Two of them are willing sitting together but only two of them never seat together.
$$\left(\begin{pmatrix}
5 \\
2
\end{pmatrix}-1\right)*4!=456$$
However the answer is 216. I don't know what's wrong.
Could you please help me find out what's wrong or give a right way to solve the problem?
Thanks!
 A: EDIT: Oh, ”onto” means it's surjective. In that case Brian M. Scott has the right answer, tough you can still solve it my way.
Old answser: Here's a tip for how I would solve it: How many functions are there from the first set to the second without restrictions? How many are there for $g(1)=g(2)$?
A: Suppose that $g$ is a function from $\{1,2,3,4,5\}$ onto $\{11,12,13,14\}$. Then exactly one of the numbers in the set $\{11,12,13,14\}$ is $g(k)$ for two different values of $k$. In terms of the children and the seats, exactly one seat has to contain two children. There are $4$ ways to choose this seat. There are $\binom52-1=9$ ways to choose which pair of children will occupy the seat. The remaining $3$ children must each take one of the other $3$ seats, and they can do this in $3!$ ways. The final result is therefore $$4\cdot\left(\binom52-1\right)\cdot3!=4\cdot9\cdot6=216\;.$$
In other words, the only thing wrong with your answer is the arithmetic:
$$\left(\binom52-1\right)\cdot4!=216\;,$$
not $456$.
(I like your translation into a problem about children and seats.)
