How many different ordered pairs of 20 elements? Question: How many different ordered pairs of 20 elements $\left( x_{1}, x_{2}, x_{3}, \dotsc, x_{20} \right)$ can you create if $x_{1}, x_{2}, x_{3}, \dotsc, x_{20}$ are non-negative integers and $x_{1} \leq 3, \, x_{2} \leq 6, x_{3} \leq 9, \dotsc, x_{20} \leq 60$ and $x_{1}, x_{2}, x_{3}, \dotsc, x_{20}$ are different?
The answer is $2^{20} \, 21!$

The number of choices that you have for the $k^{th}$ number is $3k+1-(k-1) = 2(k+1)$, because $3k+1$ is the total number of choices and $k-1$ are the numbers that have already been selected and are excluded from the available choices.
Multiplying these for $k$ ranging from $1$ to $20$ we get the desired result. Namely, $\prod \limits_{k=1}^{20} 2(k+1) = 2^{20}\cdot 21!$.
