Answer:
total no of different combinations : = 2613114
Step-by-step explanation:
Total no of different ballots that can be cast are :
City council :- (10 C 0) + (10 C 1) +(10 C 2 ) +(10 C 3)+ (10 C 4) = 446
School Board :-(8 C 0)+ (8 C 1) +(8 C 2)+(8 C 3) = 93
bond Issue: - 2^0 + 2^1 + 2^2 + 2^3+ 2^ 4 + 2^ 5 = 63
Thus total no of different combinations : = 2613114
Answer:
37,612,998 ballots
Step-by-step explanation:
The choices for city council, school board, and bond issues are independent, so we apply the product rule.
For city council, each ballot represents an unordered choice
of 0, 1, 2, 3 or 4 people from a 10 person set. Thus, there are;
C(10,0) + C(10,1) + C(10,2) + C(10,3) + C(10,4) = 1 + 10 + 45 + 120 + 210 = 386
Thus,
There are 386 many ways to fill (or partially fill) that part of the ballot.
Now, For the school board, a response is a sequence of length at most 3 from an 8 person set, with no repetitions. So, there are;
C(8,0) + C(8,1) + 2C(8,2) + 3![C(8,3)] =
1 + 8 + (2x28) + (6 x 56) = 401
Finally, each bond issue may be accepted, rejected, or left blank. Thus, this part of the ballot constitutes 5 independent choices from a three element set:
Thus, there are 3^(5) = 243 many of these. The number of different ballots that may be cast is therefore ;
386 x 401 x 243 = 37612998
