[Combinatorics]
Problem: Billy and Crystal each have a bag of 9 balls. The balls in each bag are numbered from 1 to 9. Billy and Crystal each remove one ball from their own bag. Let $b$ be the sum of the numbers on the balls remaining in Billy's bag. Let $c$ be the sum of the numbers on the balls remaining in Crystal's bag. Determine the probability that $b$ and $c$ differ by a multiple of 4 .

Solution: Suppose that Billy removes the ball numbered $x$ from his bag and that Crystal removes the ball numbered $y$ from her bag.

Then $b=1+2+3+4+5+6+7+8+9-x=45-x$.

Also, $c=1+2+3+4+5+6+7+8+9-y=45-y$.

Hence, $b-c=(45-x)-(45-y)=y-x$.

Since $1 \leq x \leq 9$ and $1 \leq y \leq 9$, then $-8 \leq y-x \leq 8$.

(This is because $y-x$ is maximized when $y$ is largest (that is, $y=9$ ) and $x$ is smallest (that is, $x=1$ ), so $y-x \leq 9-1=8$. Similarly, $y-x \geq-8$.)

Since $b-c=y-x$ is between -8 and 8 , then for it to be a multiple of $4, b-c=y-x$ can be $-8,-4,0,4$, or 8 .

Since each of Billy and Crystal chooses 1 ball from 9 balls and each ball is equally likely to be chosen, then the probability of any specific ball being chosen from one of their bags is $\frac{1}{9}$. Thus, the probability of any specific pair of balls being chosen (one from each bag) is $\frac{1}{9} \times \frac{1}{9}=\frac{1}{81}$.

Therefore, to compute the desired probability, we must count the number of pairs $(x, y)$ where $y-x$ is $-8,-4,0,4,8$, and multiply this result by $\frac{1}{81}$.

Method 1 

If $y-x=-8$, then $(x, y)$ must be $(9,1)$.

If $y-x=8$, then $(x, y)$ must be $(1,9)$.

If $y-x=-4$, then $(x, y)$ can be $(5,1),(6,2),(7,3),(8,4),(9,5)$.

If $y-x=4$, then $(x, y)$ can be $(1,5),(2,6),(3,7),(4,8),(5,9)$.

If $y-x=0$, then $(x, y)$ can be $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7),(8,8),(9,9)$.

There are thus 21 pairs $(x, y)$ that work, so the desired probability is $\frac{21}{81}=\frac{7}{27}$.

Method 2

If $x=9$, then for $y-x$ to be a multiple of $4, y$ could be 9,5 or 1 .

If $x=8$, then for $y-x$ to be a multiple of $4, y$ could be 8 or 4 .

If $x=7$, then for $y-x$ to be a multiple of $4, y$ could be 7 or 3 .

If $x=6$, then for $y-x$ to be a multiple of $4, y$ could be 6 or 2 .

If $x=5$, then for $y-x$ to be a multiple of $4, y$ could be 9,5 or 1 .

If $x=4$, then for $y-x$ to be a multiple of $4, y$ could be 8 or 4 .

If $x=3$, then for $y-x$ to be a multiple of $4, y$ could be 7 or 3 .

If $x=2$, then for $y-x$ to be a multiple of $4, y$ could be 6 or 2 .

If $x=1$, then for $y-x$ to be a multiple of $4, y$ could be 9,5 or 1 .



There are thus 21 pairs $(x, y)$ that work, so the desired probability is $\frac{21}{81}=\frac{7}{27}$.

Final answer: $\frac{7}{27}$