# Problem 5. (3 points)

In how many ways can natural numbers from 1 to 14 (each used exactly once) be arranged in a $2 \times 7$ table so that the sum of the numbers in each of the seven columns is odd?

Answer: $2^{7} \cdot(7!)^{2}$

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# Solution:

In each column, there is one even and one odd number. The choice of where to place the even number and where to place the odd number can be made in two ways, resulting in $2^{7}$ ways for the entire table. Next, there are 7! ways to arrange the even numbers in the chosen positions and the same number of ways to arrange the odd numbers.

All these numbers need to be multiplied together.