A computer will randomly generate 8 unique (no repeated numbers), two-digit numbers (10-99) and assign them a position from 1-8. I'm trying to find out the probability that the numbers will be in ascending order relative to their assigned positions. So the table generated would look something like this:
Position: 1 2 3 4 5 6 7 8
Value: a b c d e f g h
where a, b, c, ... h are random numbers from 10-99
Again, given that a-h must all be unique, two-digit numbers, what is the probability that a sequence in which a<b<c<d<e<f<g<h will occur?
Please let me know if any of this isn't clear and I'll do my best to rephrase. Thank you in advance for you help!
Because there are $8! = 40320$ possible order permutations (such as $e<g<a<c<f<b<d<h$, and so on), and each of them has equal probability to happen, so by symmetry, the probability the case $a<b<c<d<e<f<g<h$ that you want indeed happens is $1/40320$.