A 10-pack of batteries has two defective batteries. In how many ways can one select three...

Question

Question

A 10-pack of batteries has two defective batteries. In how many ways can one select three of these batteries and get

a. Neither of the defective batteries?

b. One of the defective batteries?

c. Both of the defective batteries?

Answer 1

Answer #1

a)

number of ways =N(no defective from 2 and all 3 from rest of 8) =(₈C₃)*(₂C₀) =56*1 =56

b)

number of ways =N(1 defective from 2 and all 2 from rest of 8) =(₈C₂)*(₂C₁) =28*2 =56

c)

number of ways =N(2 defective from 2 and all 1 from rest of 8) =(₈C₁)*(₂C₂) =8*1 =8

Answer 2

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