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How many arrangements of the numbers 1, 2, 3, 4, ... 7 are there where any sum of any two adjacent numbers is odd?

 Mar 6, 2023
 #1
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Notice that if we have some sequence of the numbers 1, 2, 3, 4, 5, 6, and 7, then the parity (evenness or oddness) of every odd-indexed term (the 1st, 3rd, 5th, and 7th terms) must be the same, and the parity of every even-indexed term (the 2nd, 4th, and 6th terms) must also be the same.

Let's assume without loss of generality that the 1st term is odd. Then, the 2nd term must be even, the 3rd term must be odd, the 4th term must be even, and so on.

There are 4 odd numbers and 3 even numbers, so there are 4 choices for the 1st term, and once the 1st term is chosen, there are 3 choices for the 2nd term (it must be an even number not equal to the 1st term), 3 choices for the 3rd term (it must be an odd number not equal to the 2nd term), and so on.

Therefore, the total number of arrangements is:

4 * 3 * 3 * 2 * 2 * 1 * 1 = 144

So there are 144 arrangements of the numbers 1, 2, 3, 4, 5, 6, and 7 where any sum of any two adjacent numbers is odd.

 Mar 6, 2023
 #2
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That answer is incorrect.

Guest Mar 8, 2023

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