Question

How many $2n$-digit positive integers can be formed if the digits in odd positions (counting the rightmost digit at position $1$) must be odd and the digits in even positions must be even and positive?

Answer 1

Find how many $2n$- digit positive integers can be formed:

In the $2n$-digit positive integer, there are $n$ odd positions that must be filled with digits from the set $\left\{1,3,5,7,9\right\}$ and there are $n$ even positions that must be filled with digits from the set $\left\{2,4,6,8\right\}$ .

Since $0$ neither positive nor negative so we can't include $0$ in even positive numbers.

Repetitions are allowed, and order matters.

Thus, we have

$5^n\times 4^n={\left(20\right)}^n$ possibilities where $n$ is the integer

Hence, the number of $2n$ digit positive integers with the given limitation is ${\left(20\right)}^n$.