Question #319686

consider a population consisting of 15,2,12,4,5,8,6,9 and 17.samples of size 5 are drawn from this popular

Expert's answer

C95=9!5!4!=126C^5_9=\frac{9!}{5!4!}=126 This is the number of combinations.

For example

(15,2,12,4,5)

(15,2,12,4,8)

(15,2,12,4,6)

(15,2,12,4,9)

(15,2,12,4,17)

(2,12,4,5,8)

(2,12,4,5,6)

(2,12,4,5,9)

(2,12,4,5,17)

(12,4,5,8,6)

(12,4,5,8,9)

(12,4,5,8,17)

(4,5,8,6,9)

(4,5,8,6,17)

(5,8,6,9,17)

(15,12,4,5,8)

(15,12,4,5,9)

(15,12,4,5,17)

(15,4,5,8,6)

(15,4,5,8,9)

(15,4,5,8,17)

(15,5,8,6,9)

(15,5,8,6,17)

(15,8,6,9,17)


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