P and C

Hi, there. I'm happy to help with this.

This is a really challenging question. First of all, I want to point out that this is way beyond what you would be expected to do on the GMAT. It's way beyond any of the combinatorics they ask on the AP Statistics exam. It's a hard question.

I played around with it, and from what I can tell, there's no elegant pencil-and-paper type solution. If someone can propose one, I would love to know it.

I solved this with technology. I'll explain my procedure. I used a wonderful statistics/modelling program called Fathom, which I highly recommend. On a spreadsheet in Fathom, I did the following:
(a) column #1 = the possible values for the first number, the minimum of the three -- this was simply a list from 1 to 96 (the largest possible minimum is 96, in the trio 96-98-100; any higher, and two+ of the numbers would have to be consecutive)
(c) column #2 = the allowed range for the last two --- for any given minimum, the very next number cannot be in the range. For example, if 24 is the minimum, then 25 cannot be a number in one of the allowed trios. For minimum = 24, the allowed range for the last two would be from 26 to 100, a range of 75 values. column #2 = 99 - (column #1)
(d) column #3 = the total number of pairs possible in that range (includes both allowed and not-allowed pairs). When minimum = 24, there are 75 values from which to choose for the final two, and the number of pairs one could pick is 75-C-2 (that is, "75 choose 2"). This column is calculated as: column #3 = (column #2)-C-2 = [(column #2)!]/[2!*((column #2)-2)!], where "!" is the factorial symbol.
(e) column #4 = the number of excluded pairs in that range. For example, when minimum = 24, we will pick the remaining two from the range 26 to 100, so we are guaranteed that neither will be consecutive with 24, but the two in this pair could still be consecutive with each other. The possible consecutive pairs would be (26, 27), (27, 28), . . . . (99, 100). The starting value of one of these consecutive pairs could be any value in the range except for 100, so the number of pairs of consecutive numbers in this range would be column #4 = (column #2) - 1
(f) column #5 = the number of permitted pairs, i.e. the number of non-consecutive pairs for the last two numbers. Here, column #3 is the total number of pairs in that range, and column #4 is the number that are not permitted, so I simply subtracted those. column #5 = (column #3) - (column #4)

After all that, I simply too the sum of column #5, adding up the allowed cases for each different value of the minimum number, to get the total number of allowed cases --- that is, the total number of sets of three numbers, all numbers from 1 to 100, and no two of the number consecutive. The number I got was 152096.

The total number of three-number combinations you could pick from the 100 possibilities is of course 100-C-3 = [100!]/[3!97!]=161700

Therefore, if three numbers from 1 to 100 were selected, without replacement, the probability that none of the three would be consecutive is P = 152096/161700 = 0.94060606 = 94% chance

Please let me know if you have any questions on what I've said here.

Mike