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

2016-10-20