Question 1108503
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<pre>
If three numbers from 100 to 999 are chosen at random, there is ONLY ONE permutation / arrangement to get them in increasing order 
against 6 = 3! of possible arrangements.


So, the probability under the question is {{{1/6}}}.
</pre>

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I assume (although the problem does not state it explicitly) that every time three <U>different</U> numbers are chosen at random.