Probability. Robert P. Dobrow

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rel="nofollow" href="#fb3_img_img_35014d5c-4ad4-5a09-9810-edadeae6b182.png" alt="StartSet 1 comma ellipsis comma n EndSet"/> and binary lists of length n. Table 1.3 shows the correspondence for the case n equals 3.

      A one-to-one correspondence between two finite sets means that both sets have the same number of elements. Our one-to-one correspondence shows that the number of subsets of an n-element set is equal to the number of binary lists of length n. The number of binary lists of length n is easily counted by the multiplication principle. As there are two choices for each element of the list, there are 2 Superscript n binary lists. The number of subsets of an n-element set immediately follows as 2 Superscript n.

Subset List
empty-set left-parenthesis 0 comma 0 comma 0 right-parenthesis
StartSet 1 EndSet left-parenthesis 1 comma 0 comma 0 right-parenthesis
StartSet 2 EndSet left-parenthesis 0 comma 1 comma 0 right-parenthesis
StartSet 3 EndSet left-parenthesis 0 comma 0 comma 1 right-parenthesis
StartSet 1 comma 2 EndSet left-parenthesis 1 comma 1 comma 0 right-parenthesis
StartSet 1 comma 3 EndSet left-parenthesis 1 comma 0 comma 1 right-parenthesis
StartSet 2 comma 3 EndSet left-parenthesis 0 comma 1 comma 1 right-parenthesis
StartSet 1 comma 2 comma 3 EndSet left-parenthesis 1 comma 1 comma 1 right-parenthesis

      1.7.1 Combinations and Binomial Coefficients

StartSet 1 comma 2 EndSet comma StartSet 1 comma 3 EndSet comma StartSet 1 comma 4 EndSet comma StartSet 2 comma 3 EndSet comma StartSet 2 comma 4 EndSet comma StartSet 3 comma 4 EndSet

      with corresponding lists

left-parenthesis 1 comma 1 comma 0 comma 0 right-parenthesis comma left-parenthesis 1 comma 0 comma 1 comma 0 right-parenthesis comma left-parenthesis 1 comma 0 comma 0 comma 1 right-parenthesis comma left-parenthesis 0 comma 1 comma 1 comma 0 right-parenthesis comma left-parenthesis 0 comma 1 comma 0 comma 1 right-parenthesis comma left-parenthesis 0 comma 0 comma 1 comma 1 right-parenthesis period

      Given a specific k-element subset, there are k factorial ordered lists that can be formed by permuting the elements of that subset. For instance, the three-element subset StartSet 1 comma 3 comma 4 EndSet yields the 3 factorial equals 6 lists: (1,3,4), (1,4,3), (3,1,4), (3,4,1), (4,1,3), and (4,3,1).

      It follows that the number of lists of length k made up of the elements

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