[Math] Permutation with Condition

combinatoricspermutations

The numbers 1-9 are drawn from a hat at random. Without repetition of numbers, how many numbers (with all digits drawn– eg. 987654321) would be divisible by 11?

First I recalled that a number is divisible by 11 if the number's alternating digits minus the other digits are equal to a number divisible by 11.

Honestly, I did a bit of guess and check to follow and found that 948576321 was an answer divisible by 11. Therefore if I rearranged 9,8,7,3, and 1 (the leading alternating numbers) it would bring the same result. So I solved the permutation to receive 120 possible combinations. I then took the permutation of 4,5,6, and 2 and received a total of 24 combinations.

I multiplied these two answers together to get a grand total of 2880 combinations that would be divisible by 11.

Is this the correct answer? Please help, and thanks in advance! 🙂

Best Answer

The correct answer is 31680. Your combination of alternating numbers (9, 8, 7, 3, 1) is only one out of 11 possibilities:

  1. 1, 2, 3, 4, 7
  2. 1, 2, 3, 5, 6
  3. 1, 3, 7, 8, 9
  4. 1, 4, 6, 8, 9
  5. 1, 5, 6, 7, 9
  6. 2, 3, 6, 8, 9
  7. 2, 4, 5, 8, 9
  8. 2, 4, 6, 7, 9
  9. 2, 5, 6, 7, 8
  10. 3, 4, 5, 7, 9
  11. 3, 4, 6, 7, 8

So we end up with $11 \cdot 5! \cdot 4! = 11 \cdot 120 \cdot 24 = 31680$.

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