How Many Ways Can Martians, Vesuvians, and Jovians Line Up?

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In how many ways can 5 distinct Martians, ten distinct Vesuvians and eight distinct Jovians wait together in line if no two Martains stand together>



2. 10_p_5 = 10! & 5! but what about 8? and 2?



3. 10! = 10*9*8*7*6*5*4*3*2*1
8! = 8*7*6*5*4*3*2*1
5! = 5*4*3*2*1
10*9 = 90/2 = 45?



I am so lost.. I am not a programmer..but required to take this class for my major! Please anyone help me out??
 
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consider a line of vesuvians and jovians and fill the martians into the gaps. That'll be the total possible ways of fitting in Martians so that no two matians stay together.
 
hummm

I did that one already and I didn't like the answer. But thanks.
 
RedSunflowers32 said:
I did that one already and I didn't like the answer.
what does that mean :confused:
 
It means I did the chain with every 3rd position there is a Martian, but that does not tell me how to write the equation to present the proper answer. If I factor out each element 5! = 120 * 10! = 3628800 *8! =40320
which gives me an answer of 1.5801827328E15.
Since we are working with combinations and Premutations I'm lost on which way to do the problem.
 
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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