How many arrangements of the numbers {1, 2, 3, ..., 7, 8} are there where the sum of any two adjacent numbers is odd?

\(\phantom{1, 2, 3, \dots, 7}\)

Guest Jun 19, 2022

#2**0 **

Without the repetitions of the 8 digis, you have the following permutations:

For the first 2 pairs, there are: 4 x 4 ==16 permutations. For the nest 2 pairs, there are: 3 x 3 ==9 permutations

So, the total beginning with digit will be: 16 x 9 ==144

**144 x 8 digits ==1,152 permutations**

Guest Jun 20, 2022