a computer wholesaler sells 10 different computers and each is priced differently . if the wholesaler chooses 3 computers for display at a trade show , what is the probability ( all things being equal ) that the two most expensive computers will be among the 3 chosen for display ?	Answers:	A) 15 / 56	B) 3 / 28	C) 1 / 28	D) 1 / 15	E) 1 / 168

Rationale: since , two of the choices are prefixed , we are free to choose 1 from the rest of the 8 avilable . so 8 c 1 is the numerator . total no of ways we can choose 3 from 10 is 10 c 3 , which is the denominator . so , the probability : 8 c 1 / 10 c 3 = 1 / 15 ans is d .
Answer: D