[Combinatorics]
Problem: Let $k$ be a positive integer with $k \geq 2$. Two bags each contain $k$ balls, labelled with the positive integers from 1 to $k$. André removes one ball from each bag. (In each bag, each ball is equally likely to be chosen.) Define $P(k)$ to be the probability that the product of the numbers on the two balls that he chooses is divisible by $k$.
Calculate $P(10)$.

Solution: Here, $k=10$ and so there are 10 balls in each bag.

Since there are 10 balls in each bag, there are $10 \cdot 10=100$ pairs of balls that can be chosen.

Let $a$ be the number on the first ball chosen and $b$ be the number on the second ball chosen. To determine $P(10)$, we count the number of pairs $(a, b)$ for which $a b$ is divisible by 10 .

If the number of pairs is $m$, then $P(10)=\frac{m}{100}$.

For $a b$ to be divisible by 10, at least one of $a$ and $b$ must be a multiple of 5 and at least one of $a$ and $b$ must be even.

If $a=10$ or $b=10$, then the pair $(a, b)$ gives a product $a b$ divisible by 10 .

In this case, we obtain the 19 pairs

$$
(a, b)=(1,10),(2,10), \ldots,(9,10),(10,10),(10,9), \ldots,(10,2),(10,1)
$$

If neither $a$ nor $b$ equals 10 , then either $a=5$ or $b=5$ in order for $a$ or $b$ to be divisible by 5 .

In this case, the other of $a$ and $b$ must be even and not equal to 10. (We have already counted the pairs where $a=10$ or $b=10$.)

In this case, we obtain the 8 pairs

$$
(a, b)=(5,2),(5,4),(5,6),(5,8),(2,5),(4,5),(6,5),(8,5)
$$

From our work above, there are no additional pairs for which $a b$ is divisible by 10 .

Thus, there are $19+8=27$ pairs $(a, b)$ for which $a b$ is divisible by 10 , which means that $P(10)=\frac{27}{100}$.

(We note that we could have made a 10 by 10 table that listed all possible combinations of $a$ and $b$ and their product, from which we could obtain $P(10)$.)

Final answer: $\frac{27}{100}$