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I have a question about an 8 sided die problem. I will put up my work what I have if someone can tell me how to proceed I will appreciate it.

We roll an 8 sided die numbered 1 to 8 six times and record the result of the face value that pops up on top as $x_{1}, x_{2} , x_{3}, x_{4}, x_{5}, x_{6}$.

We must now find the total number of ways that $x_{1} \ge x_{2} \cdots x_{5} \ge x_{6}$.

My progress:

If we had $6$ numbers then there will be exactly one way to arrange it such that they are going from increasing to decreasing.

Like what I mean is if the numbers that showed up were 1 through 6 then there are 720 orientations of these numbers but only one orientation will put them in increasing order.

So I thought it would be $8^6/6!$ since for every 720 combinations one puts them in increasing order, but that does not even turn out to be an integer so I am not sure what I am doing wrong.

Thanks in advance!

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    If you have $1,1,1,1,1,8$ then $120$ of the permutations satisfy the requirement rather than the one you assume.2012-12-11
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    Not sure why the dice matters here. From what you've stated, it seems like you want to count the number of feasible orderings of 6 numbers drawn from 1 to 8 *with repetition* that can be arranged in increasing order. This is simply a combinatorial question. Do you then want to ask the further question, what is the probability that six successive drawings will give you a feasible sequence?2012-12-11
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    My question is that if you roll the die 6 times you will get 6 numbers. in how many of the total possible 6 number combinations is the restraint that i gave true.2012-12-11

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