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Mathematics LibreTexts

5: Counting with Repetitions

  • Page ID
    60209
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    In counting combinations and permutations, we assumed that we were drawing from a set in which all of the elements are distinct. Of course, it is easy to come up with a scenario in which some of the elements are indistinguishable. We need to know how to count the solutions to problems like this, also.

    • 5.1: Unlimited Repetition
      For many practical purposes, even if the number of indistinguishable elements in each class is not actually infinite, we will be drawing a small enough number that we will not run out. We’ll consider two scenarios: the order in which we make the choice matters, or the order in which we make the choice doesn’t matter.
    • 5.2: Sorting a Set that Contains Repetition
      In the previous section, the new work came from looking at combinations where repetition or replacement is allowed. In this section, we’re going to consider the situation where there are a fixed number of objects in total; some of them are “repeated” (that is, indistinguishable from one another), and we want to determine how many ways they can be arranged (permuted). This can arise in a variety of situations.
    • 5.3: Summary
      This page contains the summary of the topics covered in Chapter 5.

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