Stars and Bars vs PIE I randomly made up this question so I could check: 

There are $3$ kids and $6$ gifts, how many ways to distribute so that each kid has at least one gift.

Obviously,
$**|**|**$ there are $\binom{5}{2} = 10$ solutions by stars-and-bars. 
PIE approach:
There are $3$ choices for each of the $6$ gifts $= 3^6$.
Let the kids be $ABC$, it is possible that $AB, BC, AC$ got all the gifts and one group didnt so:
$3^6 - 3(2^6)$
But then that also includes $A, B, C$ only getting the gifts (twice for each) so we add $3$.
$3^6 - 3(2^6) + 3 = 540$. 

Obviously, the first one has to be correct, but why doesnt the second approach work? 


Because in the second version you treat the gifts as different and in the first version you treat them as equivalent.
