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.
 A: You can “cast out” any $6$s from the values of the cards, since it will not affect whether the sum is a multiple of $6$. So you have 13 cards valued 1,2,3,4,5,0,1,2,3,4,4,4, and 4. How many ways can you make a multiple of 6 from the sum of 6 of these numbers? Consider the number of 4s used. If none, you can make a sum of 12 (1,2,3,1,2,3) or (0,1,1,2,3,5). With one 4, you can enumerate the possibilities, and so on. 
A: Here is a partial answer to your question.  You want the sum of two cards to be 6, 12 or 18.
For the sum of 6, the possibilities are 1+5, 2+4, and 3+3
The probability of an ace and a 5 is 1/13 times 1/12.
Same with 2+4, to wit, 1/(13*12) = 1/156.
For two threes, we have 1/13 times 3/51.
The sum of these is 2/156 + 3/(13*51)
You still need to do 12 and 18
A: Your combinations are $4\cdot 4+1\cdot 0+1\cdot 2, 2\cdot 1+2\cdot2+2\cdot3, 4 \cdot 4+2\cdot 1,+$ something that doesn't make sense because there are three places you choose from $1$ and only $0,5$ qualify.  The left number is the number of cards of that value $\pmod 6$ .  You could also have $3\cdot 4+1 \cdot 5+1 \cdot 1+1 \cdot 0, 3 \cdot 4+ 2 \cdot 2 + 1 \cdot 0$ and others
A: Answer:
Repetition of 10 is not the same as they represent different cards.  I have done it through brute force.  Hopefully I have captured everything.
Good luck
Satish
A: 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.
