How many ways are of distributing 10 balls into 4 urns such that one urn has at-least 3 balls?
My basic approach is to choose 1 urn in which I am gonna place 3 balls before hand. I can choose i urn from 4 in 4 ways.
Now I need to distribute 7 balls into 4 urns. I solve this using bars and stars. So number of ways are 10C3.
So answer is 4+10C3.
What's wrong with my approach?
It's possible for up to $3$ urns to have $3$ balls simultaneously. These will be double counted in your scheme. Also, assuming that the approach were not giving double counting, you would need to multiply the successive choices, not add them.
Incidentally, the pigeonhole principle tells us that in any distribution of $10$ balls to $4$ urns, there will always be an urn with at least $3$ balls. Stars & bars gives the distribution of $10$ indistinguishable balls to $4$ distinct urns as ${13\choose 3}=286$.