<s>[INST] 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. 
 [/INST] Case 1 is wrong because there are $4$ numbers that are neither $1$ nor $2$, so there are $6 \cdot 4^5$ ways to roll one $1$ and no $2$s.  That error repeats.  
Added:  Case 2 is wrong because the number of ways to arrange the numbers is dependent on whether there is a $2$ or not.  The $1 \cdot 1 \cdot 5 \cdot 4^3$ is really $1 \cdot 1 \cdot (4+1) \cdot 4^3$ where the $4+1$ is four ways to not get a $2$ and one way to get one.  If you don't get one there are $\frac {6!}{2!4!}={6 \choose 2}=15$ ways to arrange them.
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