Find the sum of all $abc$ 
Let $T$ be the set of all triplets $(a,b,c)$ of integers such that $1 \leq a < b < c \leq 6$. For each triplet $(a,b,c)$ in $T$ , take number $a \cdot b \cdot c$. Add all these numbers corresponding to all the triplets in $T$. Prove that the answer is divisible by $7$.

Attempt
Let $S = \displaystyle \sum_{b < c} bc $. Then with no restriction on $a$ we have $M = (1+2+3+4+5+6)S = 21S$. This is the sum of the cases where $a < b, a = b, $ and $a > b$. Now how can I use this to find the case where $a<b$?
 A: For any such triplet $(a, b, c)$ with $a<b<c$, 
we have a corresponding triplet $(7-c, 7-b, 7-a)$ with $7-c<7-b<7-a$.
Since we can pair up all the triplets in this manner, 
and $abc+(7-c)(7-b)(7-a)\equiv 0\pmod{7}$,
the sum of the products of each of the triplets is divisible by 7.
A: A triple $(a, b, c)$ of integers such that $1 \leq a < b < c \leq 6$ is one of the $\binom{6}{3} = 20$ three-element subsets of the set $\{1, 2, 3, 4, 5, 6\}$ because once we choose a subset there is only one way of placing the elements in increasing order.  They are 
\begin{array}{c c}
(a, b, c) & abc\\
(1, 2, 3) & 6\\
(1, 2, 4) & 8\\
(1, 2, 5) & 10\\
(1, 2, 6) & 12\\
(1, 3, 4) & 12\\
(1, 3, 5) & 15\\
(1, 3, 6) & 18\\
(1, 4, 5) & 20\\
(1, 4, 6) & 24\\
(1, 5, 6) & 30\\
(2, 3, 4) & 24\\
(2, 3, 5) & 30\\
(2, 3, 6) & 36\\
(2, 4, 5) & 40\\
(2, 4, 6) & 48\\
(2, 5, 6) & 60\\
(3, 4, 5) & 60\\
(3, 4, 6) & 72\\
(3, 5, 6) & 90\\
(4, 5, 6) & 120
\end{array}
As you can check, the sum of the products is divisible by $7$.
