# 2013 AMC 12A Problems/Problem 20

## Problem 20

Let $S$ be the set $\{1,2,3,...,19\}$. For $a,b \in S$, define $a \succ b$ to mean that either $0 < a - b \le 9$ or $b - a > 9$. How many ordered triples $(x,y,z)$ of elements of $S$ have the property that $x \succ y$, $y \succ z$, and $z \succ x$? $\textbf{(A)} \ 810 \qquad \textbf{(B)} \ 855 \qquad \textbf{(C)} \ 900 \qquad \textbf{(D)} \ 950 \qquad \textbf{(E)} \ 988$

## Solution 1

Imagine that the 19 numbers are just 19 persons sitting evenly around a circle $C$; each of them is facing to the center.

One may check that $x \succ y$ if and only if $y$ is one of the 9 persons on the left of $x$, and $y \succ x$ if and only if $y$ is one of the 9 persons on the right of $x$. Therefore, " $x \succ y$ and $y \succ z$ and $z \succ x$" implies that $x, y, z$ cuts the circumference of $C$ into three arcs, each of which has no more than $10$ numbers sitting on it (inclusive).

We count the complement: where the cut generated by $(x,y,z)$ has ONE arc that has more than $10$ persons sitting on. Note that there can only be one such arc because there are only $19$ persons in total.

Suppose the number of persons on the longest arc is $k>10$. Then two places of $x,y,z$ are just chosen from the two end-points of the arc, and there are $19-k$ possible places for the third person. Once the three places of $x,y,z$ are chosen, there are three possible ways to put $x,y,z$ into them clockwise. Also, note that for any $k>10$, there are $19$ ways to choose an arc of length $k$. Therefore the total number of ways (of the complement) is $$\sum_{k=11}^{18} 3\cdot 19 \cdot (19-k) = 3\cdot 19 \cdot (1+\cdots+8) = 3\cdot 19\cdot 36$$ $$3\cdot \binom{19}{3} - 3\cdot 19\cdot 36 = 3\cdot 19 \cdot (51 - 36) = 855$$

NOTE: this multiple-choice problem can be done even faster -- after we realized the fact that each choice of the three places of $x,y,z$ corresponds to $3$ possible ways to put them in, and that each arc of length $k>10$ has $19$ equitable positions, it is evident that the answer should be divisible by $3\cdot 19$, which can only be $855$ from the five choices.

## Solution 2

First, we can find out that the only $x, y, z$ that satisfy the conditions in the problem are $(1 \leq z-y \leq 9 , 1 \leq y-x \leq 9 , z-x \geq 10)$ , $(1 \leq y-x \leq 9 , 1 \leq x-z \leq 9 , y-z \geq 10)$ , and $(1 \leq x-z \leq 9 , 1 \leq z-y \leq 9 , x-y \geq 10)$.

Consider the 1st set of conditions for $x, y, z$. We get that there are $$45\cdot 10 - \sum_{k=1}^{9} \binom{k+1}{2} = 285$$

cases for the first set of conditions.

Since the 2nd and 3rd set of conditions are simply rotations of the 1st set, the total number of cases is $$3\cdot 285 = \fbox{(B)855}$$

## Solution 3 (Common sense)

The conditions are very strange. The major way $x,y,z$ can satisfy is if $(1 \leq z-y \leq 9 , 1 \leq y-x \leq 9 , z-x \geq 10)$ because at this point $z > x$ no matter what. This is the only case that works, but we can rotate the variables around, so our answer is a multiple of $3$.

Suppose $z=y+a$ and $y=x+b$ where $0. That means $z=x+a+b$ and $10 \leq z \leq 18$. WLOG $z>y>x$. Start casework on what $a+b$ is, anywhere from $10$ to $18$. There are 2 main factors: 1. the number of ways to choose $a,b$, which is $a+b-1$. 2. the number of ways to choose $x$ or start the sequence, which is $19-(a+b)$ The sum of all cases turns out to be the sum of squares from $9^2$ to $1^2$, which is $\frac{9 \cdot 10 \cdot 19}{2 \cdot 3}$. Multiply that by $3$ and get $\fbox{(B)855}$

~dragnin

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