In a group of 8 engineers, 3 engineers have a PHD degree; the others have only an MS or BS degree. A team of 3 engineers is to be chosen which must have at least 1 engineer with a PHD, how many different teams can be chosen?  Choose the correct choice
A) 56 B) 46 C) 64 D) 65 E) 10

Rationale: the problem asks for a combination , since order does n ' t matter . now , selecting r items from a set of n gives the combination formula : ncr = n ! / r ! ( n - r ) ! n = 8 r = 3 so , total teams is 8 c 3 = 8 ! / ( 3 ! ( 8 - 3 ) ! ) = 56 , and n = 8 - 3 = 5 r = 3 for teams without a phd is 5 c 3 = 5 ! / ( 3 ! ( 5 - 3 ) ! ) = 10 , so , teams with at least 1 phd = 56 - 10 = 46 answer : b
Answer: B