I am a little confused at how to solve the Permutations

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SUMMARY

The problem involves calculating the number of ways to arrange five distinct Martians, ten distinct Vesuvians, and eight distinct Jovians in a line with the condition that no two Martians stand together. The correct approach involves first arranging the 18 Vesuvians and Jovians, which can be done in 18! ways. This creates 19 potential slots for the Martians. The final calculation for the arrangement is 18! * P(19,5), resulting in 1,395,360 valid arrangements where the Martians are placed in the available slots without violating the condition.

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Bucs44
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I am a little confused at how to solve the following problem:

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


What is throwing me off is the addition of another group - if it was just Martians and Vesuvians it would be easy. My guess is P(10,5) = (10*9*8*7*6)

Am I correct or did I mess this up?
 
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It doesn't matter how the Vesuvians and Jovians stand so thinking of these as ALIENS you DO have two groups 5 Martians and 18 ALIENS
 
So my calculation was wrong? It should be P(18,5) ?
 
How many ways could you arrange the ALIENS? Remember they are all distinct.

How many ways could you then slot in the Martians to meet the conditions?
 
Right - Because there are 18 Vesuvians and Jovians, the Martians can either fit in the front of the line, middle or end of the line. In which case there would be 19 spots where they could fit in line. P(19,5) = 1,395,360
 
Bucs44 said:
Right - Because there are 18 Vesuvians and Jovians, the Martians can either fit in the front of the line, middle or end of the line. In which case there would be 19 spots where they could fit in line. P(19,5) = 1,395,360

Is this right?
 

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