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In a standard deck, there are 13 cards in each suit,such as hearts. For simplicity we'll number them from 1 to 13. This question is about these 13 cards.

Start with the cards in the usual order. This time you are allowed to divide the deck into three piles, consisting of the top n cards, the next m cards, and the remaining 13-n-m cards. Keeping the order within these piles unchanged, you place one pile at the top, one in the middle, and one on the bottom. You may repeat this as many times as you like, changing m and n if you wish. How many arrangements can you get?

My answer:

So we can number them from 1 to 13 in the usual order: 1 2 3 4 5 6 7 8 9 10 11 12 13

So as one case I can have firs pile=4 cards, second pile=4 cards and last pile=5 cards. If I do the described permutation, I get:

1 5 9 2 6 10 3 7 11 4 8 12 13

Is this correct? If this is correct, then this question is asking me to consider different values of m and n and consider different arrangements based on these values?

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    So are you allowed to shuffle the piles in any order? but when the question said "Keeping the order within these piles unchanged, you place one pile at the top, one in the middle, and one on the bottom.". I thought that it means shuffling piles in order so you place top pile at top, middle pile at the middle and bottom pile at the bottom.2011-03-08

3 Answers 3

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Try to think of a generating set of $S_{13}$ that you can implement using this operation.

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    Remember, you can perform the operation an arbitrary number of time with arbitrary $n$ and $m$.2011-03-08
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To put the card in position $n$ at the top without changing the order of the other cards, split the deck into $(n-1)$, $1$ and $13-n$ piles and restack as $1$, $(n-1)$ and $(13-n)$. You don't need to do this if $n=1$, while if $n=13$ then split into something like $10$, $2$ and $1$ and restack as $1$, $10$ and $2$.

Do this up to 13 times and you can have any order.

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    @Henry: Thanks again. just the last question, with your answer, how should I do that up to 13 times? I do not understand how I must do that! Sorry if my question is dumb but I have thought about this for a week and I still with your answer do not know how I should do it. If you show just a few steps of what you do, I would be grateful to you.2011-03-09
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Every permutation of a finite set can be expressed as the product of transpositions.

Obviously, under the system I have described, I can transpose the top two cards. but I do not know how to devise a method of moving any two cards to the top without disturbing the other cards?

It is unclear whether m or n are allowed to be zero. Can my method deal with the situation if they are not?