## ParthKohli one year ago If a nine-digit number $$x$$ is formed using 1, 2, 3, 4, 5, 6, 7, 8, 9 without repetition, then find the probability that the non-negative difference among all the digits equidistant from both ends is 1.

1. ParthKohli

The central number has to be odd. 5 ways to choose an odd number. Now if 3 is chosen then the fixed pairs are (1, 2), (4, 5), (6, 7), (8, 9) so 4! ways to move the pairs around and $$2^4$$ ways to flip within the pair.$\frac{4!\times 5 \times 2^4}{9!}= \frac{1}{189}$Not a part of the choices...

2. thomas5267

How does this work? 864,152,579 is one such number. But 5 is equidistant to both ends and 5-5=0.

3. ParthKohli

Nah, the number at the center is not considered, because that way, there would be no such number.

4. FireKat97

5. ganeshie8

Just to confirm that your answer is correct, I am getting the same 1920 numbers that satisfy the given requirement by bruteforce : https://jsfiddle.net/ganeshie8/Lqbagygh/embedded/result/

6. ParthKohli

Thank you sir. Please post that here. http://math.stackexchange.com/questions/1469932/the-probability-that-non-negative-difference-of-the-digits-at-equal-distances-fr

7. ParthKohli

Oh, that question has an answer that exactly matches mine. Good.

8. ganeshie8

I don't see 1/189 in that question...

9. ParthKohli

Exactly, it doesn't match the choices.

10. ganeshie8

I'm not gonna spend more time on this.. typoes/gross mistakes are typical in textbooks written by indian authors in rush

11. ParthKohli

12. ganeshie8

"non- negative difference" are we interpreting this particular phrase correctly ?

13. ParthKohli

I guess we are.

14. ParthKohli

$|a-b| = 1$

15. thomas5267

Non-negative difference does not make much sense since the order of the subtraction is not specified.