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

1.
consider the case n=5,

2n=10 so we have a 10 digit number where the 1, 3, 5, 7, 9th positions are odd and 2, 4, 6, 8, 10th are even.

For example 2 3 2 3 2 3 2 3 2 3 is such a number.

2.
There are 5 odd positions, each of which can be any of {1, 3, 5, 7, 9}

and 5  even positions, any of which can be any of {2, 4, 6,8 }.

In total we can form 5*5*5*5*5*4*4*4*4*4=$5^{5}*4^{5}=20^{5}$ such numbers.

3. Similarly in a 2n digit integer, there are n positions, each with 5 possible choices to fill with, and n positions, each with 4 possibilities to fill.

This means that there are in total $5^{n}*4^{n}=20^{n}$ numbers that can be formed under the conditions given.

Answer: $20^{n}$