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.

Question

parthkohli · Answer

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...

thomas5267 · Answer

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

parthkohli · Answer

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

firekat97 · Answer

what are your options?

ganeshie8 · Answer

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/

parthkohli · Answer

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

parthkohli · Answer

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

ganeshie8 · Answer

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

parthkohli · Answer

Exactly, it doesn't match the choices.

ganeshie8 · Answer

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

parthkohli · Answer

Ahahaha, why that username

ganeshie8 · Answer

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

parthkohli · Answer

I guess we are.

parthkohli · Answer

$$|a-b| = 1$$

thomas5267 · Answer

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