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Since I can vote for one, or two, or three, or more candidates, provided I indicate my preferences, how many different ways could ballots be filled out when there are 10 candidates? Print E-mail

With just two candidates you could vote 4 different ways: C1,C2,C1C2, and C2C1, and with three candidates you could vote 15 different ways:
C1,C2,C3, C1C2,C1C3,C2C1,C2C3,C3C1,C3C2,C1C2C3,C1C3C2,C2C1C3,C2C3C1,C3C1C2, and C3C2C1.  In mathematical terms, if there are n candidates, the number of different ways ballots could occur is nP1+nP2+nP3+...+nPn, where nPm is the number of permutations of n objects chosen m at a time.  As you can see from the table below, the number of distinct ballot permutations grows very rapidly as the number of candidates increases.  

# Candidates                              # Distinct Ballots

1                   1
2                   4
3                   15
4                   64
5                  325
6                 1,956
7                 13,699
8                109,600
9                986,409
10               9,864,100
11              108,505,111
12             1,302,061,344
13             16,926,797,485
14            236,975,164,804
15           3,554,627,472,075
16           56,874,039,553,216
17          966,858,672,404,689
18         17,403,456,103,284,420
19        330,665,665,962,403,999
20       6,613,313,319,248,080,000

So, the answer to your question is that with 10 candidates there are more than three times as many ways to fill out a ballot than there are registered voters in BC.  With 13 candidates there would be more than twice as many ways to fill out a ballot than there are people on Earth.