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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:

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$\textbf{Case 1:}$ 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.

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  • $\begingroup$ You can write $\cdot$ and $\times$ to produce, respectively, the multiplication symbols $\cdot$ and $\times$. $\endgroup$ – N. F. Taussig Oct 16 '18 at 8:17
  • $\begingroup$ In this question triple is treated as three pairs or no pair? $\endgroup$ – Jaroslaw Matlak Oct 16 '18 at 11:45
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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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  • $\begingroup$ Thank you for catching that error, would you mind checking if other than that the response is correct? $\endgroup$ – Mathaholic24 Oct 16 '18 at 5:03
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    $\begingroup$ @Mathaholic24 In Cases 3 and 4, you also have to take into account how many times each digit appears in the sequence when accounting for the number of distinguishable sequences. $\endgroup$ – N. F. Taussig Oct 16 '18 at 9:05
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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
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