Sum of 6 cards being multiple of 6 I pick 6 cards from a set of 13 (ace-king). If ace = 1 and jack,queen,king = 10 what is the probability of the sum of the cards being a multiple of 6? 
Tried so far:
I split the numbers into sets with values:
6n, 6n+1, 6n+2, 6n+3
like so:
{6}{1,7}{2,8}{3,9}{4,10,j,q,k}{5}
and then grouped the combinations that added to a multiple of 6:
(5c4)(1c1)(2c1) + (2c2)(2c2)(2c2) + (5c4)(2c2) + (5c2)(1c1)(1c1)(2c1)(2c1)
/ (13c6)
= 10/1716
I am almost certain I am missing combinations but am having trouble finding out which.

It will help to explicitly list the values of cards in each combination, as other answers have done, then count the number of ways you can make each combination.
If doing this by hand rather than by computer, you were wise to group the cards by
their remainders after division by $6$; you can use just the remainders as the terms
of your sum (as already shown in two answers).
For example, you can make a multiple of $6$ by taking $4+4+4+4+5+3=24.$
There are exactly $\binom54\binom11\binom21$ ways to draw cards without replacement 
whose remainders would add up in exactly this way. 
(You could also write this expression $(5C4)(1C1)(2C1)$ as you did in the question.)
But you can also make a multiple of $6$ by taking $4+4+4+4+2+0=24.$
There are also exactly $\binom54\binom11\binom21$ ways to draw such a set of cards
(or you could write $\binom54\binom21\binom11$ so that the terms you are multiplying
are in the same order as the corresponding terms of the sum, for clarity),
and these sets are all different from the ones that are counted in the previous paragraph.
You might make a table of sums and the number of ways to draw each sum:
$$\begin{array}{lcrcr}
4+4+4+4+5+3=24 && \binom54\binom11\binom21&=&10 \\
4+4+4+4+2+0=24 && \binom54\binom21\binom11&=&10 \\
4+4+4+4+1+1=24 && \binom54\binom22&=&5 \\
4+4+4+5+1+0=18 && \binom53\binom11\binom21\binom11&=&20
\end{array}$$
Continue this column for several more rows until you have listed all the sums that
are possible to draw that are divisible by $6.$
Then you can add up the values in the right-hand column.
