ParthKohli
  • ParthKohli
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.
Mathematics
chestercat
  • chestercat
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ParthKohli
  • 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...
thomas5267
  • thomas5267
How does this work? 864,152,579 is one such number. But 5 is equidistant to both ends and 5-5=0.
ParthKohli
  • ParthKohli
Nah, the number at the center is not considered, because that way, there would be no such number.

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FireKat97
  • FireKat97
what are your options?
ganeshie8
  • 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/
ParthKohli
  • 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
ParthKohli
  • ParthKohli
Oh, that question has an answer that exactly matches mine. Good.
ganeshie8
  • ganeshie8
I don't see 1/189 in that question...
ParthKohli
  • ParthKohli
Exactly, it doesn't match the choices.
ganeshie8
  • ganeshie8
I'm not gonna spend more time on this.. typoes/gross mistakes are typical in textbooks written by indian authors in rush
ParthKohli
  • ParthKohli
Ahahaha, why that username
ganeshie8
  • ganeshie8
"non- negative difference" are we interpreting this particular phrase correctly ?
ParthKohli
  • ParthKohli
I guess we are.
ParthKohli
  • ParthKohli
\[|a-b| = 1\]
thomas5267
  • thomas5267
Non-negative difference does not make much sense since the order of the subtraction is not specified.

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