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Answer to Question #349391 in Statistics and Probability for hiro

Question #349391 
For N= 20 and n = 18, how many possible combinations of samples without replacement can be formed?





1
Expert's answer
2022-06-09T14:50:38-0400
"\\dbinom{N}{n}=\\dbinom{20}{18}=\\dfrac{20!}{18!(20-18)!}=190"


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