Was just thinking if my solution is correct.
I have 60 kids, and 20 benches with 4 seats each - 80 different seats.
While order in each bench is important, how many ways to place kids when only one of the benches is fully seated (say bench number i)?
My solution is to first place the 4 kids (to first seat in bench i has 60 kids to choose, for second seat 59 and so on) so thats: $\ P^{60}_4$
And for the rest of the benches - 19 (they all have to be not full) not really sure what to do, thought $\ P^{60-4}_{80-4}$ but this doesn't really seem to be right...
Edit: actually first expression should be $P^4_{60}$ I believe, since that for the bench $i$ it's first seat has 60 kids to choose from, the second seat has 59 and so on..
As easy to see there are one bench fully seated, one seated with 2 kids, and 18 seated with 3 kids, so that one obtains the following number of ways to place kids: $$ \frac {20!}{1!1!18!}\times\frac {60!}{4!2! (3!)^{18}}\times\left [\frac {4!}{(1!)^40!}\frac{4!}{(1!)^22!}\left (\frac{4!}{(1!)^31!}\right)^{18}\right], $$ where the first factor counts the number of ways to permute one 4-seated bench, one 2-seated and 18 3-seated ones, the second factor counts the number of ways to spread the kids among the benches, and the third factor counts the number of ways to arrange the kids on each bench (including the indistinguishable empty spaces).