What is the probability of getting 1,2,3,4 when 5 dice are thrown and

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What is the probability of getting 1,2,3,4 when 5 dice are thrown and one of them is not 5.(the order of 1,2,3,4 doesn't matter)
I am really confused:
So I computed chances of getting correct numbers that is 1/6^4 times 5/6,then I multiply this by 4 times 5! Am I right?
 
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MilenaMath said:
What is the probability of getting 1,2,3,4 when 5 dice are thrown and one of them is not 5.(the order of 1,2,3,4 doesn't matter)
Your question is unclear. Do you mean at least one each of 1, 2, 3, 4? Exactly one of each?
When you say one of them is not five, do you mean that a particular one is not five, that they are not all five, or that none of them are five?
 


I need to get exactly 1,2,3,4(order doesn't matter) and the fifth dice can be whatever number but 5
 


Sorry, still not clear. If you get 4, 2, 2, 3, 1, is that ok or not?
 


MilenaMath said:
I need to get exactly 1,2,3,4(order doesn't matter) and the fifth dice can be whatever number but 5

From this description, it seems the event you want is
A = {1,1,2,3,4} or {1,2,2,3,4} or {1,2,3,3,4} or {1,2,3,4,4} or {1,2,3,4,6}, but without regard for order. Is that correct? If so, you can use a multinomial distribution to find
P{A} = p(2,1,1,1,0,0) + p(1,2,1,1,0,0) + ... + p(1,1,1,1,0,1), where these are the occupancy numbers in each of the 6 classes 1--6.

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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