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# lec0127 - Some Counting Last lecture I ended with an...

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Some Counting... Last lecture I ended with an example where you could determine the size of a particular set given the size of other sets, by using a Venn diagram. In this lecture, I will formalize some of these ideas. (Give you formulas pertaining to how to count various properties of sets.) First of all, probably the easiest rule is the summation rule. If you have two sets A and B with A B = , then we have: |A B| = |A| + |B| So, the size of the union of two disjoint sets is the sum of the size of the two sets. (For example, given a set of 8 girls and 9 boys, there is a total of 17 kids in the union of the two sets.) Before I get to the next rule, let me define a Cartesian product. This is defined as follows: A X B = { (a,b) | a A b B } This is a set of ordered pairs, hence the order here matters. The next rule is the product rule. If we want the size of the Cartesian Product of two sets, we can get it as follows: |A X B| = |A|x|B| This can generalize into answering questions such as how many possible binary strings of length 7 are there?

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Combinations Now, using the tools that we have we can figure out the answer to the particular problem: How many subsets of a particular size(say, k) does a set of size n have. (0 k n) Luckily, there is a special mathematical symbol to denote the number of ways of choosing k distinct objects out of n distinct objects. This is called a combination or a binomial coefficient. In particular, you can do this in exactly n C k ways. Now, the next question is, what number does that equal? Here it is: n C k = n!/(k!(n-k)!), where n! = 1x2x3x4...xn This certainly does not look like an obvious result. And it is not. So, let me go ahead and prove it to you.
Consider you have a set of numbers {1,2,3,...,n} and you want

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