# Problem 5. (3 points)

In how many ways can natural numbers from 1 to 10 (each used exactly once) be arranged in a $2 \times 5$ table so that the sum of the numbers in each of the five columns is odd?
Answer: $2^{2} \cdot(5!)^{2}$

Solution:

In each column, there is one even and one odd number. Where the even number stands, and where the odd number stands, can be chosen in two ways, making a total of $2^{5}$ ways for the entire table. Next, there are 5! ways to arrange the even numbers in the chosen places and the same number of ways to arrange the odd numbers.

All these numbers need to be multiplied together.