How many different ways can a deck of cards (52 facedown) be stacked?

Question

While playing a card game with my nephew, he asked about shuffling the cards.

Let's say I am playing a card game and the stack of cards is shuffled between each game. (note - the cards must all be face down)

What is the total number of different ways the cards can be stacked?

I wrote on piece of paper 52! and explained the multiplication to him and plugged it into the iPhone calculator. The answer seemed to be a bit large, so I decided to open a question here.

Thanks

Answer 1

The way you have to see this is the following.

First you should look at wheater this is a combination, or a permutation. The usual way to do this is to ask "Does Order matter?" (i.e you have to ask yourself if having a queen of diamonds at the bottom of your stack is the same as having a king of hearts at the bottom of your stack ? Are these two stacks the same ? The answer is no, they're not the same). Thus, this is a permutation.

Second, you should ask yourself: "Is repetition allowed ?" (i.e can you have, for example, two queen of hearts in your stack? Again, you cannot, so repetition is not allowed).

The formula for permutation without repetition (a.k.a permutation without replacement) is the following: $$P(n,r) = \frac{n!}{(n-r)!}$$ where $n$ is the number of cards you can pick (you have a total of $52$ cards), and $r$ is the number of cards you are picking (and stacking).

In your case, you have $n=52$ cards and you are stacking $r=52$ cards (you're stacking all of them), so you have: $$P(52,52) = \frac{52!}{0!}\\=52!$$ (note that $0! = 1$)

Thus the answer to your question is $52!$ (assuming your cards cannot be upside down).

Now, say that instead of stacking all the cards, you want to know how many stacks of $7$ cards you can make with one deck (i.e a total of $52$ cards), then the answer is simply $P(52,7)$.

Answer 2

Total number of cards = 52

Total number of ways = 52!

For first position, we have 52 choices, for second position we have 51 choices and so on, $52*51*.. *2*1 = 52!$

Answer 3

$52$ choices for the first card, then $51$ for the second (because $1$ card has been fixed), then $50$ for the 3rd (Because $2$ cards have been fixed)..... etc 1 for the 52nd because 51 have been fixed. Thus at each of the 52 stages, we have chosen 1 card. By the Rule of product, we multiply the number of choices at each stage giving $52\cdot51\cdot50\cdot \cdots \cdot 1 = 52!$

Answer 4

The number of ways are $52!$ As for I place you have 52 cards, for II 51.... and so on.

Basically it is permutation of 52 things of which no two are same.

Therefore 52! I hope you asked for same.

Answer 5

$2^{52} * 52!$, because some of the cards might be upside down