COT3100Counting02 - Variations of counting combinations...

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Variations of counting combinations Combining our result for counting combinations, some logic, the sum rule and the product rule, we can handle more sophisticated counting questions. Take the following, for example: Let S= {1, 2, 3, . .., 30}. How many subsets A of S contain 5 elements, with 5 being the least? In essence, we know that 5 must be one of our elements, so we are really free to choose only 4 elements. But, we have a restriction here too. We must choose those four elements from the set {6, 7, . .., 30}. The number of ways to do this are 25 C 4 . How many subsets A of S contain 5 elements with the smallest element not being equal to 5? We know the total number of subsets of S that are of size 5 is are 30 C 5 . And we also know that of these, exactly are 25 C 4 have 5 as the smallest element. Thus, our answer should be the difference of these two, or 30 C 5 - 25 C 4 .
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How many subsets of S contain 5 elements with the smallest element less than 5? This is actually quite a difficult question. The problem is that we don’t know how many of the elements are less than 5. In fact we have 4 (disjoint) possibilities: 1 element is less than 5: Thus we choose 1 element from the set {1,2,3,4} and 4 elements from the set {5,6,. ..,30}. Since these choices are independent, we can invoke the product rule to find the total number of ways to do this as 4 C 1 * 26 C 4 . 2 elements are less than 5: Thus we choose 2 elements from the set {1,2,3,4} and 3 elements from the set {5,6,. ..,30}. Since these choices are independent, we can invoke the product rule to find the total number of ways to do this as 4 C 2 * 26 C 3 . 3 elements are less than 5: Thus we choose 3 elements from the set {1,2,3,4} and 2 elements from the set {5,6,. ..,30}. Since these choices are independent, we can invoke the product rule to find the total number of ways to do this as 4 C 3 * 26 C 2 . 4 elements are less than 5: Thus we choose 4 elements from the
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COT3100Counting02 - Variations of counting combinations...

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