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I have a set $A=\{0,1,2,3,4,5,6,7,8\}$. I construct numbers of three digits from the set $A$ and I am not allowed to repeat a number. I put them in a new set:

$B=\{012,013,014,\dots,081,082,083,\dots,107,108,120,123,124,\dots,874,875,876\}$.

I want to know the probability that a number is less than $368$. Supposedly the answer is $\dfrac{209}{504}$, however, I found my answer by counting each case (the hardest way) and it is $\dfrac{230}{560}$. So, if anyone has a better method in order to find the correct answer, I'll appreciate it.

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    There are $504$ numbers, of which $368$ is the $210$th, counting from least to greatest, so I think $209/504$ is the correct answer. You must have double-counted some numbers, or included some invalid numbers, perhaps? – 2017-02-02

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The denominator $= 9\cdot 8\cdot 7 = 504$

The numerator $\{0,1,2\}$ followed by anything two digits $= 3\cdot 8\cdot 7$

$+ 3$ followed by $\{0,1,2,4,5\}$ flowed by what ever is available $= 1\cdot 5\cdot 7$

$+ 3$ followed by $6$ followed by $\{0,1,2,4,5,7\} = 6$