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# What fraction of all the 10-digit numbers with distinct digits have the property that the sum of every pair of neighboring digits is odd?

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What fraction of all the \$10\$-digit numbers with distinct digits have the property that the sum of every pair of neighboring digits is odd?

My thought process:

Ok, so distinct digits means that there is only one of each digit. So all digits are used.

The only two forms of the numbers in terms of even and odd is "eoeoeoeoeo" or "oeoeoeoeoe", with e for even and o for odd.

For eoeoeoeoeo, there's 4 choices for the first, 5 for the second, 4 for the third, 4 for the fourth, 3 for the fifth, etc. I get 4*5*4*4*3*3*2*2, or 11520.

For oeoeoeoeoe, I get 5*5*4*4*3*3*2*2, or 14400.

Adding 11520 and 14400 up, I get 25920. There are 9 billion 10-digit numbers in total. 25,920/9,000,000,000 is my fraction, which simplifies to 9/3,125,000

But that's incorrect. What's the correct way? And if possible, to whoever answers this, point out my errors?

(also PLEASE, PLEASE don't just put one answer. Even just showing your calculations would be okay. Don't bother wasting your time or energy on this problem if you can't explain. It will prevent others from answering this problem, since one answer would already be there.)

Thanks!

Mar 31, 2021