Bunuel wrote:
A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?
(A) 6
(B) 8
(C) 9
(D) 12
(E)16
English is one of the six but must be included. We really have five courses and need to pick three of them. That's\(\frac{5!}{3!2!} =\frac{5*4}{2} =10\) . That's all the possibilities, so we need to exclude the disallowed cases.
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