if there are 4 different tables and 18 people in total, how many ways can the people be seated to have at least 4 in each table?

Question

There are 2 possible arrangements: 4,4,4,6 and 4,4,5,5

For, 4,4,4,6: divide 18 people in these groups: 18C4⋅14C4⋅10C4⋅6C6 seat them in: 3!⋅3!⋅3!⋅5!

(note that you dont have a choice in selecting a table, since all 4 are alike)

For, 4,4,5,5: divide 18 people in these groups: 18C4⋅14C4⋅10C5⋅5C5 seat them in: 3!⋅3!⋅4!⋅4!

What would be the final solution statement?