Japan 2013 Initial Round

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 $$\left(\genfrac{}{}{0ex}{}{26}{2}\right)=\frac{26\times 25}{2\times 1}=325.$$ 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 $$1\times 1+32\times 3+32\times 3+2304\times 6=14017.$$