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### Combinations without replacement

The only difference between this case and the case of permutations without replacement is that the selection order of the k selected objects is not distinguished here. This implies that a different arrangement of the same objects is not counted for this case and so this case involves simply selecting a subset of size k from the population. Therefore, we can view the number of permutations without replacement as a two-stage process: first select a subset (combinations without replacement) and then generate every possible rearrangement of each of these subsets. Note that the number of ways to generate every possible rearrangement of k objects is equivalent to counting the number of permutations without replacement of k objects selected from a population of size k, and so is equal to k!. Denote by C(n,k) the number of combinations without replacement. Then we have,

Hence,

This quantity is usually denoted by

Larry Ammann
2017-04-26