A certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 3 of 10 candidates eligible to fill 3 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 4 candidates are there to fill the 4 positions?  Pick one:
A) 420
B) 770
C) 840
D) 165
E) 315
1 of 7 will be chosen for the math   3 of 10 will be chosen for the computer   None of the 4 chosen people can be in more than one departments.   We can choose any of the 7 candidates for the math dep., which gives as 7 selections.   We can choose 3 of the 10 candidates for the computer dep., which gives us 3 selections and 7 rejections.   So, the way to find how many different selections of 2 candidates we can have for the computer dep., we do:   10! /3!*7!= 120   We are multiplying our individual selections:  7*120= 
<gadget id="calculator"> 7*120</gadget>
<output>840</output>
840   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 840
<result>C</result>
A certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 3 of 10 candidates eligible to fill 3 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 4 candidates are there to fill the 4 positions?