You have 8 identical candies and 4 identical cookies to give to 3 children (two boys and a girl). In how many ways can this be done if (a) each child gets at least one of each type, (b) each child gets exactly four items, (c) the girl gets more candies than either boy and the boys get equal numbers of cookies?

Step-by-step solution

1. 1

   Set up the tool. The number of ways to split $n$ identical items among $k$ distinct children with each getting at least $1$ is $\binom{n-1}{k-1}$; if zero is allowed it is $\binom{n+k-1}{k-1}$. Candies and cookies are distributed independently, so in each part the two counts multiply.

2. 2

   Part (a): each child gets at least one of each. For the candies, $\binom{8-1}{2}=\binom72=21$; for the cookies, $\binom{4-1}{2}=\binom32=3$. Multiplying,

   $21\times 3=63.$

3. 3

   Part (b): each child gets exactly four items. Here the two distributions are no longer independent — the cookie count determines the candy count. If child $i$ gets $k_i$ cookies then it gets $4-k_i$ candies, and $\sum k_i=4$ automatically forces $\sum(4-k_i)=12-4=8$ candies. So it is enough to count cookie splits with zeros allowed:

   $\binom{4+2}{2}=\binom62=15.$

4. 4

   Part (c), cookies: the boys must get equal numbers. If each boy gets $k$ cookies then the girl gets $4-2k$, which needs $k\in\{0,1,2\}$ — three possibilities.

5. 5

   Part (c), candies: the girl beats both boys. Enumerate by the girl's share $g$, with the boys' shares summing to $8-g$ and each below $g$:

   $g=4:3,\quad g=5:4,\quad g=6:3,\quad g=7:2,\quad g=8:1,$

   and $g\le 3$ is impossible since then the boys would need $\ge 5$ candies between them with each below $3$. That totals $3+4+3+2+1=13$ candy distributions.

6. 6

   Combine for part (c) and summarise.

   $13\times 3=39.$

   So the three answers are $\mathbf{63}$, $\mathbf{15}$ and $\mathbf{39}$. A direct computer enumeration over all distributions reproduces each of these counts exactly.