Unfortunately, we approach directly by brute force. I do not see a more convenient approach. We break first into cases by candybar distribution, and then break down further based on lolipop distribution.
Either the person with six candybars does or does not get two lollipops: 2 options
Choose whether the person with 5 candybars does or does not get two lollipops and choose whether the person with 2 candybars does or does not get two lollipops. $4$ options
Similar to last case: $4$ options
Choose whether the person with 4 candybars does or does not get two lollipops. Then, choose whether one, both, or none of the persons with 2 candybars get two lollipops. Then, note that it is impossible for all three of them not to get two lollipops, so subtract one from the count to correct that: $2\cdot 3-1=5$ options
Similar to last case, choose for the person with 2 candybars, then choose how many people with 3 candybars for $5$ options.
Choose whether the person with 3 candybars does or does not get two lollipops. Then, choose whether the person with 1 candybar does or does not get two lollipops. $4$ options
Clearly only one option here.
We have then for a grand total $2+4+4+5+5+4+1=25$ options
As you observed, there are $12^{20}$ ways to distribute the presents without restriction. There are $\binom{12}{k}$ ways to exclude $k$ of the recipients from receiving a present and $(12 - k)^{20}$ ways to distribute the presents to the remaining $12 - k$ people. By the Inclusion-Exclusion Principle, the number of ways to distribute the presents so that each person receives at least one is
$$\sum_{k = 0}^{12} (-1)^k\binom{12}{k}(12 - k)^{20}$$
Best Answer
We can distribute $3$ teddy bears to $4$ children in $4^3$ ways.
We can ensure that each child gets $3$ goodies by filling out each child with lollipops.
So, there are a total of $4^3=64$ ways