Question 383202: I don't see my catagory. It is about permutations. My book is not giving me clear answers. I need to find the number of permutations for: 1,3,5? and another is: 7P3? Thank you.
Found 2 solutions by jim_thompson5910, moncherie1:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! The set of permutations of {1,3,5} is simply the set of all possible ways of arranging {1,3,5}. These are
{1,3,5}, {1,5,3}, {3,1,5}, {3,5,1}, {5,1,3}, {5,3,1}
Since there are 6 items in this collection, this means that there are 6 permutations.
A quick way of computing this is to simply calculate 3! to get 3! = 3*2*1 = 6
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Recall that
n P r = (n!)/(n-r)!
So in this case, n = 7 and r = 3, which means that
7 P 3 = (7!)/(7-3)! = (7*6*5*4!)/(4!) = 7*6*5 = 210
So 7 P 3 = 210
Answer by moncherie1(1) (Show Source):
You can put this solution on YOUR website! 8.62 In 1992, the FAA conducted 86,991 pre-employment drug tests on job applicants who were to be
engaged in safety and security-related jobs, and found that 1,143 were positive. (a) Construct a
95 percent confidence interval for the population proportion of positive drug tests. (b) Why is the
normality assumption not a problem, despite the very small value of p? (Data are from Flying 120,
no. 11 [November 1993], p. 31.)
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