Given a train with 20 different carts with 4 seats each, how many options are there for 60 people to sit, while order in each cart matters, and there is no full cart?
Using inclusion-exclusion:
$W(0) = \binom{80}{60}*60!*20!=80!$ (choosing 60 seats, rearranging people, rearranging carts)
$P_i$ - cart $i$ is full
now for every $0\leq r \leq 15$ we get $W(r)=\binom{20}{r}\frac{60!}{(60-4r)!}$ and for $16 \leq r \leq 20$ $W(r) = 0$ because we can't have 16 full carts.
so the solution is $80! + \sum_{i=1}^{15}(-1)^r \binom{20}{r}\frac{60!}{(60-4r)!}$
does this seem right?
I would not have interpreted the problem as allowing the carts to be rearranged, but either way, I see no need for an inclusion-exclusion argument. Imagine that the carts are lined up. We can line up the people in $60!$ orders. No cart is full, so each cart must get exactly $3$ people. We put the first $3$ people in the first cart, the next $3$ people in the second cart, and so on. In each cart we seat the $3$ people in order in the first $3$ seats.
However, we’ve counted only those arrangements that use the first $3$ seats of each cart. A given group of $3$ people can actually choose seats in a cart in $4\cdot3\cdot2=24$ ways, and our count includes only $3!=6$ of those ways. Thus, in each of the $20$ carts there are really $4$ times as many arrangements as we’ve counted, and therefore we need to multiply our first estimate of $60!$ by $4^{20}$ to get $4^{20}\cdot60!$.
If the carts are indistinguishable, there is definitely no need to take into account their order: they are identified only by their actual position in the queue. If they do have individual identities, so that two arrangements in which the same people are in the same seats in the $k$-th cart for $k=1,\ldots,20$, but the carts themselves have been permuted so that different carts in some positions, then this must be further multiplied by $20!$.