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Starting with a standard $52$ deck of cards how many ways can you order all $52$ cards. But for this, the actual suits don't matter. So if you have a particular order, then you switch the position of the $2$ of clubs with the $2$ of hearts, that is considered the same set.

user118494
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Socko
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2 Answers2

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There are $52!$ ways of ordering a deck of cards normally.

Consider that each of 13 ranks has 4 identical cards, and remove their internal permutations,

thus $\dfrac{52!}{{(4!)}^{13}}$

[ Just as permutations of $AABB = \dfrac{4!}{(2!)^2}$ ]

true blue anil
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  1. In how many ways can you order $52$ cards in the regular case where suits matter? Let's call the number of ways $C$.
  2. Say the ace cards lose suit in the $C$ different orderings. This creates groups of orderings which are now indistinguishable - cases where ace cards occupy the same positions in the deck, with different positions of suits. How big is one such group? Let's call the size of such a group $p$.
  3. What is the number $\frac{C}{p}$ a count of?
  4. Say now also the king cards lose suit...
milcak
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  • This means that C = 52! But I don't know how to figure out what p would be. – Socko Sep 22 '15 at 19:19
  • $p$ is the number of ways you can order $4$ jokers of different suits into $4$ slots - their positions in the given card ordering. – milcak Sep 22 '15 at 20:02
  • I don't know what that is, or how to figure it out. – Socko Sep 22 '15 at 20:11
  • How many jokers can you put into first slot? * How many can you now put into second? * Third? * Forth? --- same idea as getting $C$. – milcak Sep 22 '15 at 20:13
  • It doesn't make sense. There are 52 cards (no jokers), and each set of 4 cards could be in any of those 52 slots. – Socko Sep 22 '15 at 21:57
  • Sorry about that - forget the joker consider aces instead... There are $4$ aces in each ordering of cards, $4$ different suits can be put in the first slot, $3$ suits in second, $2$ in third, and $1$ in last slot that remain. Hence $p = 4!$. So $52!/4!$ will be the number of orderings of the $52$ cards in which the suit of aces doesn't matter. – milcak Sep 23 '15 at 00:14