A ballot lists ten candidates for city council, eight candidates for the school board, and five bond issues. The ballot instructs voters to choose up to four people running for city council, rank up to three candidates for the school board, and approve or reject each bond issue. How Many different ballots may be cast, if partially completed (or empty) ballots are allowed

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

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