Question

Among 15 clocks there are two defectives. In how many ways can an inspector chose

three of the clocks for inspection so that:

a. There is no restriction.

b. None of the defective clock is included.

c. Only one of the defective clocks is included.

d. Two of the defective clock is included.

Answer 1

a.$C^3_{15}=\frac{15!}{12!3!}=455$

b.$C^3_{15-2}=\frac{13!}{10!3!}=286$

c. $C^1_2C^2_{13}=\frac{2!}{1!1!}\frac{13!}{11!2!}=156$

d.$C^2_2C^1_{13}=1\cdot\frac{13!}{12!1!}=13$