Given $6$ distinct objects and $10$ distinct bins how many ways are there to distribute the objects such that at least one bin contains more than one object? What if the objects are identical?
If the objects are distinct I think there's a total of: $$ {10-1+6\choose 6}6! $$ ways to distribute the objects. We can calculate in how many ways we can distribute the objects such that each cell contains exactly $0$ or $1$ objects: $$ {10\choose 6}6! $$ We choose $6$ bins out of $10$ that will contain an object and we permute them. So the final answer is: $$ {10-1+6\choose 6}6!-{10\choose 6}6! $$
If for the same question the objects are identical then we don't need to permute the objects so the answer is: $$ {10-1+6\choose 6}-{10\choose 6} $$
I'm really not sure if the solution is correct.
There are $10$ bins in which each of the six objects could be placed, so there are $10^6$ ways to distribute the objects without restriction.
The restriction that at least one bin contains more than one object means that the six objects cannot be placed in six different bins. In how many ways could this occur?
There are $\binom{10}{6}$ ways to choose six bins to receive the objects and $6!$ ways to distribute six distinct objects to those bins. Hence, the number of ways the six distinct objects could be distributed so that each object is placed in a separate bin is $$P(10, 6) = \binom{10}{6}6! = \frac{10!}{4!6!} \cdot 6! = \frac{10!}{4!}$$ Hence, the number of ways to distribute six distinct objects to ten distinct bins so that at least one bin contains more than one object is $$10^6 - \binom{10}{6}6!$$
You answered this question correctly.