<s>[INST] Three terms from a sequence Find the number of ways of choosing three numbers from the set ${1,2,3,…,20}$, so that the sum of the three numbers is divisible by $3$. I wrote $a+b+c=3k$, where $k=1,2,...20$, till $k=6$, there were no problems, but then restrictions started($a<21$). Even if I come to know of the correct method from here, the summation of all the answers is a tedious job(the answers come in binomial coefficients). Therefore, s there a shorter way?
 [/INST] We only need concern ourselves with congruence $\bmod 3$.
There are $7$ numbers that are $1\bmod 3$
There are $7$ numbers that are $2\bmod 3$
There are $6$ numbers that are $0\bmod 3$
We now count the ways to add to $0\bmod 3$:

Three zeros:
$0+0+0$ (There are $\binom{6}{3}$ of these)

Two zeros:

One zero:
$0+1+2$ (There are $7\cdot7\cdot 6$ of these )

No zeros:
$1+1+1$ (There are $\binom{7}{3}$ of these)
$2+2+2$ (There are $\binom{7}{3}$ of these)
Adding up the answer is $\binom{6}{3}+7\cdot7\cdot6+2\binom{7}{3}=384$
There are $\binom{20}{3}=1140$ ways to choose, $3$ numbers of $20$. A third of that would be $380$. So more than a third of the subsets of size three are $0\bmod 3$
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