Combinatorics rolling a die six times How many ways are there to roll a die six times such that there are more ones than twos?
I broke this up into six cases: 
$\textbf{EDITED!!!!!}$
$\textbf{Case 1:}$ One 1 and NO 2s --> 1x4x4x4x4x4 = $4^5$. This can be arranged in six ways: $\dfrac{6!}{5!}$. So there are $\dfrac{6!}{5!}$$4^5$ ways for this case. 
$\textbf{Case 2:}$ Two 1s and One 2 OR NO 2s --> 1x1x5x4x4x4 = $5x4^3$. This can be arranged in $\dfrac{6!}{2!3!}$ ways. So there are $(6x4^3)$$\dfrac{6!}{2!3!}$ ways for this case. 
$\textbf{Case 3:}$ Three 1s and Two, One or NO 2s --> 1x1x1x5x5x4 = 4x$5^2$. This can be arranged in $\dfrac{6!}{2!3!}$ ways. So there are $(4x5^2)$$\dfrac{6!}{2!3!}$ ways for this case. 
$\textbf{Case 4:}$ Four 1s and Two, One or NO 2s --> 1x1x1x1x5x5 = $5^2$. This can be arranged in $\dfrac{6!}{4!2!}$ ways. So there are $(5^2)$$\dfrac{6!}{4!2!}$ ways for this case. 
$\textbf{Case 5:}$ Five 1s and One or NO 2s --> 1x1x1x1x1x5 = 5. This can be arranged in $\dfrac{6!}{5!}$ ways = 6 ways. So there are $5^3$ ways for this case. 
$\textbf{Case 6:}$ Six 1s and NO 2s --> 1x1x1x1x1x1 = 1. There is only one way to arrange this so there is only 1 way for this case. 
With this logic...I would add the number of ways from each case to get my answer. 

If the question is "where is my error", than plese ignore this answer, which is giving an other way to count. The idea is that there are either more $1$'s, case (1), or more $2$'s, case (2), or they occur equaly often, case $(=)$. Of course, the count of possibilities for case (1) is the same as for case (2), so we simply count the possibilities for $(=)$, subtract them from the total, divide by $2$, so the number we are searching for is:
$$
N = \frac 12\left(\ 6^6 
-\binom 60\binom 60 4^{6-0}
-\binom 61\binom 51 4^{6-2}
-\binom 62\binom 42 4^{6-4}
-\binom 63\binom 33 4^{6-6}
\ \right) \ .
$$
The products of binomial coefficients of the shape $\binom 6k\binom{6-k}k$, $k=0,1,2,3$ count the possibilities to fix the $k$ places among $6$ places for the $1$'s, then the $k$ places for the $2$'s from the remaining $6-k$.
We get then
$$
N = \frac 12\left(\ 6^6 
- 4^6 
- 6\cdot 5\cdot 4^4 
- 15\cdot 6\cdot 4^2
- 20\cdot 1\cdot 4^0
\ \right) 
=
16710\ .
$$
We can also check this answer with sage, by enumerating all possibilities.
sage: len( [ x for x in cartesian_product( [[1..6] for _ in [1..6]] )
....:        if list(x).count(1) > list(x).count(2) ] )
16710

