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Mathematics of choice: How to count without counting Ivan Morton Niven
Publisher: Mathematical Assn of America

This voting practice will not be and each had to be hand-counted. There must be two boys together, and they Or else we could slip $2$ boys into one of the two center gaps ($2$ choices), and then slip the remaining boy into one of the $3$ remaining gaps, for a total of $6$ choices. From an online course or community college course.” Betsy: Geometric Shapes and Beginning Fractions “Ordering a pizza became a math lesson, not only in counting the money to pay the bill but also for fractions problems. After all, even the person most allergic to math, most traumatized by math, still remembers how to count! Since we have already counted the number of "bad" positions with all the boys together, it remains to count the number of bad positions in which the boys are not all together, but some boy is not next to a girl. That's because we all fall prey to the belief that we can have our own side conversations that are quiet enough not to disrupt the counting – unlike those other loudmouths. Boltzmann's farout idea is now causing cosmologists fits, If they calculate the chance that the cosmological constant will taken on any particular value in a randomly chosen universe, then the value of our cosmological constant will probably be one of the popular choices. As you see, this “counting” is a little more challenging than the kind of “counting” you learned in your salad days. This odd scenario is really, really, really unlikely — but not, quite, impossible, as physicist Ludwig Boltzmann realised in the nineteenth century. For example: Shakespeare wrote fifteen comedies and ten histories. This year, a voter who votes for the same candidate for all three choices, called repeat candidate voting, will give that candidate one first-choice vote but will not have the second and third choices counted. The registration requirement will still allow voters to cast their ballot for Mickey Mouse or Donald Duck, but there will be no count beyond the total number of write-in votes for non-registered write-ins. If option #1 has P alternatives and option #2 has Q alternatives (assuming that the two sets of alternatives have no overlap), then total number of different pairs we can form is P*Q.