# Cardinality of permutations definition

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?

• you mean "cardinality of the set of permutations"
– YCor
Mar 24, 2019 at 17:42

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.

• 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! Mar 24, 2019 at 15:39
• "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. Mar 24, 2019 at 15:51
• Thank you! It helped me a-lot! Mar 24, 2019 at 16:45