# Number of surjective functions$f: A->B$ where $f(1) > f(2) > f(3)$

"Find the number of surjective functions $f: A->B$, where$A = \{1,2,3,4,5,6\}, B = \{1,2,3,4\}$ and where $f(1) > f(2) > f(3)$."

I know how to find # of surj. functions from $A->B$ (in this case $4^6$minus all maps to subsets of B). But I don't know how to implement this condition$f(1) > f(2) > f(3)$.

I think it is important to calculate maps of $\{1, 2, 3\}$, I have found 4 ways how to do the mapping, simply by writing all the possible ways (I think it is 4 choose 3 since there is only one way how to order). Then I need to assign some $(>=1)$ elements from $\{4, 5, 6\}$ to the remaining element of $B$ (for the function to be surjective) and keep in mind that $\{4, 5, 6\}$ can also be assigned to other els of$B$.

Can you please give me an advice how to do it? Thank you

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You're on the right track. Now just observe that the number of ways to map $4$, $5$ and $6$ such that the missing element is covered is the total number of ways to map them minus the number of ways to map them to one fewer element.