# 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.

• You can write $\cdot$ and $\times$ to produce, respectively, the multiplication symbols $\cdot$ and $\times$. – N. F. Taussig Oct 16 '18 at 8:17
• In this question triple is treated as three pairs or no pair? – Jaroslaw Matlak Oct 16 '18 at 11:45

## 2 Answers

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.

• Thank you for catching that error, would you mind checking if other than that the response is correct? – Mathaholic24 Oct 16 '18 at 5:03
• @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. – N. F. Taussig Oct 16 '18 at 9:05

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