Fifty identical chocolate bars are to be distributed among $8$ children. If the oldest three must have a total of $10$ bars and the youngest three must have at least $3$ each. In how many ways can the bars be distributed?
Oldest 3 children need 10 bars: $\binom{10+3-1}{3} = \binom{12}{3}= 220 $
Youngest 3 children need 9 or more (3 each) = $\binom{9+3-1}{3} = 165$
$31$ bars are left to distribute to the youngest children and the remaining 2 children: $\binom{31+5-1}{5} = 324,632$
So there would be a total of $325017$ ways to distribute with the above restrictions.
You are reading the problem to say the oldest three must share exactly $10$, not at least $10$, which makes sense to me. For the younger three you can just give them three each and distribute the other $41$. The ten to the oldest three can be distributed in ${10+3-1\choose 3-1}=66$ ways. We should multiply that by the number of ways to distribute the other $31$ among five people, which is ${31+5-1 \choose 5-1}=52360$ The final count is $66 \cdot 52360=3455760$