, 2013

Let us denote by $a_X$, $a_Y$ and $a_Z$ the numbers of cakes of type $A$ received by $X$, $Y$ and $Z$ respectively, and by $b_X$, $b_Y$ and $b_Z$ the numbers of cakes of type $B$ received by $X$, $Y$ and $Z$, respectively. Then, they are non-negative integers and they satisfy $a_X + a_Y + a_Z = b_X + b_Y + b_Z = 24$.

If $a_X < a_Y$, then by the condition specified for the problem, we must have either $b_X > b_Y$ or $a_X + b_X \ge a_Y + b_Y$. We note that even when the second alternative takes place, we must have $b_X > b_Y$, since $a_X < a_Y$. So, we must have the following implication: $\lceil a_X < a_Y \rceil \Rightarrow \lceil b_X > b_Y \rceil$. Arguing similarly, we see that in order for the condition of the problem to be satisfied, we must also have the following implications as well: $\lceil a_X > a_Y \rceil \Rightarrow \lceil b_X < b_Y \rceil$, $\lceil b_X < b_Y \rceil \Rightarrow \lceil a_X > a_Y \rceil$, $\lceil b_X > b_Y \rceil \Rightarrow \lceil a_X < a_Y \rceil$. We therefore conclude that in order for the condition of the problem to be satisfied for $X$ and $Y$, we need one of the following conditions to be satisfied:

* $a_X < a_Y$ and $b_X > b_Y$,
* $a_X = a_Y$ and $b_X = b_Y$,
* $a_X > a_Y$ and $b_X < b_Y$.

Conversely, if one of these conditions is satisfied, then we see the condition of the problem is satisfied for $X$ and $Y$. Similar statements can be made for $X$ and $Z$, and for $Y$ and $Z$, which guarantee the validity of the condition of the problem for the corresponding pairs.

The number of triples $(x, y, z)$ of non-negative integers satisfying $x + y + z = 24$ is given by
$$\binom{26}{2} = \frac{26 \times 25}{2 \times 1} = 325.$$
We classify them further according to the relative order of $x, y, z$.

* When $x = y = z$ is satisfied: there is only one triple $(x, y, z) = (8, 8, 8)$ in this case.
* When $x = y < z$ is satisfied: in this case, we can write $(x, y, z) = (k, k, 24-2k)$, where $k$ is an integer satisfying $0 \le k \le 7$. So, there are $8$ such triples. The same result holds for the cases $y = z < x$, $z = x < y$.
* When $x = y > z$ is satisfied: in this case, we have $(x, y, z) = (k, k, 24-2k)$, where $k$ is an integer satisfying $9 \le k \le 12$. So, there are $4$ such triples. The same result holds for the cases $y = z > x$, $z = x > y$.
* The remaining case: We have $x, y, z$ to be distinct in this case, and there are $325-1-8 \times 3-4 \times 3 = 288$ such triples $(x, y, z)$. There are $6$ possibilities for the order of $x, y, z$, but by symmetry, we can conclude that the number of those triples $(x, y, z)$ with $x < y < z$ is $\frac{288}{6} = 48$, and the same result holds for others.

In order to get the solution for the problem, we count the number of cases depending on the order relation among $a_X, a_Y, a_Z$.

* When $a_X = a_Y = a_Z$: in this case, we must have $b_X = b_Y = b_Z$, so we have only $1 \times 1 = 1$ possibility.
* When $a_X = a_Y < a_Z$: in this case, we have to have $b_X = b_Y > b_Z$, so we have $8 \times 4 = 32$ possibilities. The same result holds for the cases $a_Y = a_Z < a_X$ and $a_Z = a_X < a_Y$.
* When $a_X = a_Y > a_Z$: in this case, we have to have $b_X = b_Y < b_Z$, so we have $4 \times 8 = 32$ possibilities. The same result holds for the cases $a_Y = a_Z > a_X$ and $a_Z = a_X > a_Y$.
* When $a_X < a_Y < a_Z$: in this case, we have to have $b_X > b_Y > b_Z$, so we have $48 \times 48 = 2304$ possibilities. The same result holds for the five other cases, where $a_X, a_Y, a_Z$ are all distinct.

Summing up all these numbers of possibilities, we obtain that the number of ways to distribute cakes among $X$, $Y$, $Z$ to satisfy the condition of the problem is
$$1 \times 1 + 32 \times 3 + 32 \times 3 + 2304 \times 6 = 14017.$$