I'm studying the chapter of counting for my discrete math exam and I'm getting a bit confused with the terms permutations/combinations, elements and ways.

As far as I know:

  • Permutations/combinations are an arrangement of ordered/unordered distinct elements.
  • Elements are what permutations/combinations are made of.
  • When we refer to the number of permutations/combinations of an arrangement, we mean how many ways we can do a permutation/combination of that arrangement.

But then there are also the terms r-permutation and r-combination where $r$ are the elements of an arrangement. So, when we do a permutation $P(n,r)$, what is $n$ then? I thought $n$ was the number of elements of our set.

Here's one of the references I used:

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I'm a little bit confused... Could someone clarify this for me?

Thank you in advance.

  • $\begingroup$ I would hope that the book provided definitions and examples for these terms. Can you edit your question to post a particular paragraph or statement that puzzles you? $\endgroup$ Commented Dec 9, 2017 at 18:56
  • $\begingroup$ $R$ should be the number of elements that are permuted and $N$ should be the total number of elements. For example, $P(5, 3)$ means $\frac{5\cdot 4\cdot 3\cdot 2\cdot 1}{2\cdot 1}$ which is basically the number of ways to pick 3 things out of 5 where order does matter. $\endgroup$
    – NL628
    Commented Dec 9, 2017 at 19:04

1 Answer 1


Here, as an example, $$P(5, 3) = \frac{5!}{2!}$$ means that out of 5 possible objects, how many ways there are to choose 3 objects where order matters. On the other hand, $$C(5, 3) = \frac{5!}{3!\cdot 2!}$$ means the same thing except order doesn't matter.

In this case, note that the textbook says "where order matters." Hence, $P(N, R)$ is number of ways to choose R objects where order matters and $C(N, R)$ is the number of ways to choose an object where order doesn't matter.

Does this help?

  • $\begingroup$ So it means that we have a set of 5 elements, say {a,b,c,d,e}, you change 3 elements, hence you do a 3-permutation or a 3-combination. If you care about the order, it means that you choose the three first elements of the set (permutation). And if you don't care about the order, then you choose them randomly (combination). Is that right? $\endgroup$
    – Arnau
    Commented Dec 9, 2017 at 19:19
  • $\begingroup$ @Arnau Yes, that is correct. Did my answer help? $\endgroup$
    – NL628
    Commented Dec 9, 2017 at 19:23

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