The formula for the following conditions is [tex]C(n,r) = \dfrac{n!}{[(n-r)!r!]}[/tex]
How many ways k things out of m different things (m ≥ k) can be chosen if order of the chosen things doesn't matter?
We can use combinations for this case,
A total number of distinguishable things is m.
Out of those m things, k things are to be chosen such that their order doesn't matter.
This can be done in a total of
[tex]^mC_k = \dfrac{m!}{k! \times (m-k)!} ways.[/tex]
If the order matters, then each of those choice of k distinct items would be permuted k! times.
So, a total number of choices, in that case, would be:
[tex]^mP_k = k! \times ^mC_k = k! \times \dfrac{m!}{k! \times (m-k)!} = \dfrac{m!}{ (m-k)!}\\\\^mP_k = \dfrac{m!}{ (m-k)!}[/tex]
This is called the permutation of k items chosen out of m items (all distinct).
The formula for the following conditions is [tex]C(n,r) = \dfrac{n!}{[(n-r)!r!]}[/tex]
Learn more about combinations and permutations here:
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