<s>[INST] Probability Problem on Divisibility of Sum by 3 From the 3-element subsets of $\{1, 2, 3, \ldots , 100\}$ (the set of the first 100 positive integers), a subset $(x, y, z)$ is picked randomly. What is the probability that $x + y + z$ is divisible by 3?
This is a math Olympiad problem.  I would welcome a good solution.
 [/INST] Note that our set has $34$ members equivalent to $1$ modulo $3$ (that is, with remainder $1$ upon division by $3$), $33$ equivalent to $2$, and $33$ equivalent to $0$.
Suppose we draw three numbers at random to create our subset. In order to get a sum divisible by $3$, we could draw any of the following sets of remainders:
$$
2,2,2\\
1,1,1\\
0,0,0\\
0,1,2 \text{ (In each of }3! \text{ arrangements)}
$$
So, the number of possible drawings that will yield a multiple of $3$ (order matters here) is 
$$
(34\times 33\times 32)+ (33\times 32\times 31)\times 2+(33\times 34 \times 33)\times (3\times 2 \times 1)
$$ 
There are $100\times 99\times 98$ possible ways of drawing three numbers, our probability is
$$
\frac{(34\times 33\times 32)+ (33\times 32\times 31)\times 2+(33\times 34 \times 33)\times (3\times 2 \times 1)}{100\times 99 \times 98}
$$
Which, upon simplification, yields an answer of $\frac{817}{2450}$.

It is also possible to approach this with combinations instead of permutations.  That is, noting we must have either all in one category or one from each category, our total number of possibilities is
$$
\binom{33}{3} + \binom{33}{3} + \binom{34}{3} + 33\times 34 \times 33
$$
producing a probability of
$$
\frac{\binom{33}{3} + \binom{33}{3} + \binom{34}{3} + 33\times 34 \times 33}{\binom{100}{3}}
$$
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