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?  Select the correct option.
A) 56	B) 46	C) 64	D) 65	E) 10
The problem asks for a combination, since order doesn'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 8C3 = 8!/(3!(8 - 3)!) = 56, and   n = 8 - 3= 
<gadget id="calculator">8 - 3</gadget>
<output>5</output>
5   r= 3   for teams without a PHD is 5C3 = 5!/(3!(5 - 3)!) = 10,   so, teams with at least 1 PHD = 56 - 10= 
<gadget id="calculator">56 - 10</gadget>
<output>46</output>
46   Answer: 46
<result>B</result>
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?