# Number of surjections from 8-element set to 6-element set

My attempt:

Surjections from 8-element set to 6-element can be divided into 2 cases.

1. Three of the elements from 8-element set are mapped to single element in 6-element set.

2. Four of the elements from 8-element set, which are divided into groups of two, are mapped to two elements in 6-element set.

For the first case, we have $${8}\choose{3}$$ $$\cdot$$ $$6!$$ possible surjections.

The second case has $${8}\choose{2}$$ $$\cdot$$ $${6}\choose{2}$$ $$\cdot$$ $$6!$$ possible surjections.

In total, there are $${8}\choose{3}$$ $$\cdot$$ $$6!$$ + $${8}\choose{2}$$ $$\cdot$$ $${6}\choose{2}$$ $$\cdot$$ $$6!$$ $$= 342720$$ possible surjections.

Is this a correct approach to solve the problem?

Almost perfect. You simply have to apply the $$(1/2)$$ scalar to the second case to compensate for over-counting.
For example: If you are sending $$\{a,b,c,d,e,f,g,h\}$$ to $$\{1,2,3,4,5,6\}$$
your present formula counts twice $$(a,b) \to 1, ~~(c,d) \to 2$$.