In a group of 12 engineers, 4 engineers have a PHD degree; the others have only an MS or BS degree. A team of 4 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) 495  B) 70  C) 245  D) 425  E) 555
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 = 12   r = 4   So, total teams is 12C4 = 12!/(4!(12 - 4)!) = 495, and   n = 12 - 4= 
<gadget id="calculator">12 - 4</gadget>
<output>8</output>
8   r= 4   for teams without a PH425 is 8C4 = 8!/(4!(8 - 4)!) = 70,   so, teams with at least 1 PH425 = 495 - 70= 
<gadget id="calculator">495 - 70</gadget>
<output>425</output>
425   Answer: 425
<result>D</result>
In a group of 12 engineers, 4 engineers have a PHD degree; the others have only an MS or BS degree. A team of 4 engineers is to be chosen which must have at least 1 engineer with a PHD, how many different teams can be chosen?