I'm trying to understand the definition of cardinality of permutations through a basic example. If, for example, there is a set:

$A = \{2, 3, 4, 2, 1\}$

What is the cardinality of its permutations?

And now if I add another two sets:

$B = \{7, 8, 9\}\\ C = \{1,2\}$

Can I define the cardinality of permutations between the three sets?

  • 2
    $\begingroup$ you mean "cardinality of the set of permutations" $\endgroup$
    – YCor
    Mar 24, 2019 at 17:42

1 Answer 1


Your vocabulary is not quite right but I think I know what you mean.

What you call the "cardinality of permutations" of $2,3,4,2,1$ is the number of different ways to write those digits in some order. One such way is $23421$, another is $12234$.

If all five digits were different then there would be $5! = 120$ ways. But think about the two $2$'s. Imagine for the moment that one of them is red and the other black. Then you could tell the difference between the two ways to write $23421$ depending on where the red $2$ appeared. So if you answered $120$ you would be counting those two as different when they are really the same. You have double counted. The correct answer is $60 = 5!/2$.

This logic generalizes. If there were three $2$'s you would count each arrangement $3! = 6$ times. So for example putting $A$ and $C$ together your list is $1,1,2,2,2,3,4$ and there are $7!/2!3!$ ways to arrange it.

Last note: you should not write $A$ as you have and call it a set since sets can't have repeated elements. It's a list, or a multiset.

  • $\begingroup$ Thank you very much!! I think I understand the first part. Regarding the second part, if I have 3 sets (proper ones with distinct elements), can I define permutations between them? Thanks again! $\endgroup$ Mar 24, 2019 at 15:39
  • $\begingroup$ "Permutations between three sets" doesn't make sense to me. You can put those three lists (not sets) together to get $1,1,2,2,2,3,4,7,8,9$ which can be rearranged (permuted) $10!/2!3!$ ways. If you have three genuine sets with no duplications in their union then the permutation count will be just $n!$ when there are $n$ distinct elements. $\endgroup$ Mar 24, 2019 at 15:51
  • $\begingroup$ Thank you! It helped me a-lot! $\endgroup$ Mar 24, 2019 at 16:45

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