What you want to know is that if we are given two lists of numbers $$1, \dots, m$$ and $$1, \dots, n$$
Then how many ways are there to pick two numbers $(a, b)$ where $a$ is in the first list and $b$ is in the second list.
Note that if we fix the number $1$ from the first list. Then we can pair that number up with $n$ different numbers. More specifically, we would have the pairs $$(1, 1), \dots, (1, n)$$
This tells us that if fix a number $k$ from the first list, then we get $n$ different pairs $$(k, 1),\dots, (k, n)$$
Because there are $m$ numbers in the first list. And each number will give us $n$ pairs. The total number of pairs is $$\underbrace{n+\dots+n}_\text{$m$ times}=m\times n$$
In your specific example, the two lists are $$1, 2, 3, \dots, 10$$ and $$1, 2, 3, \dots, 10$$
Thus, the total number of ways to pick a pair $(a, b)$ where $a$ comes from the first list and $b$ comes from the second list is $10\times 10=100$.
This result holds in general. This is known as the principle of multiplication or the rule of product. Thus, if you have $n$ lists $$1, \dots, m_1\\ \vdots \\ 1,\dots ,m_n$$
Then the total number of ways to pick $n$ numbers where each number comes from each list is $$m_1\times \cdots \times m_n$$
Hope this helps!