A certain university will select 1 of 6 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department. If none of the candidates is eligible for a position in both departments, how many different sets of 3 candidates are there to fill the 3 positions?
Choose the correct choice from the following answers.
A) 42  B) 270  C) 140  D) 165  E) 315
1 of 6 will be chosen for the math   2 of 10 will be chosen for the computer   None of the 3 chosen people can be in more than one departments.   We can choose any of the 6 candidates for the math dep., which gives as 6 selections.   We can choose 2 of the 10 candidates for the computer dep., which gives us 2 selections and 8 rejections.   So, the way to find how many different selections of 2 candidates we can have for the computer dep., we do:   10! /2!*8!= (9*10) / 2 = 90 / 2= 
<gadget id="calculator">90 / 2</gadget>
<output>45</output>
45.   We are multiplying our individual selections:  6*45= 
<gadget id="calculator"> 6*45</gadget>
<output>270</output>
270   In the bolded part, we don't have to multiply all of the numbers, as those in 8! are included in 10!, so we simplify instead.   Ans 270
<result>B</result>
A certain university will select 1 of 6 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department. If none of the candidates is eligible for a position in both departments, how many different sets of 3 candidates are there to fill the 3 positions?