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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.
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