Question 808939: I have 60 patients that will be in blocks of 6. There will be 3 treatments available, A , B, C. I am having difficulty listing all of the possible permutations. I could really use some help. Thank you!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! I have 60 patients that will be in blocks of 6. There will be 3 treatments available, A , B, C. I am having difficulty listing all of the possible permutations.
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There are 60C3 = (60*59*58)/(1*2*3) = 34,220 unique sets of 3 you
could possibly have.
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Assuming you mean to assign A,B, and C randomly to the 3 persons
in a set, there are 3! = 6 ways to make the assignments.
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Cheers,
Stan H.
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