Combinations

Sutured

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Jan 2, 2014
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Hello all. It is my first time in this forum.

English is not my native language so please bear with me.

I have a standard deck of cards and I take 6 of them. The order doesn't matter so I can get them with:

52! / 6! * (52-6)! = 20358520 combinations.

Now, assuming that all aces have the same value, all twos the same ... all kings the same, how many combinations are there?

For example the combination:
1 of spades, 2 of spades, 3 of spades, 4 of spades, 5 of spades, 6 of spades, is the same as
1 of clubs, 2 of spades, 3 of spades, 4 of spades, 5 of spades, 6 of spades, and
1 of clubs, 2 of spades, 3 of spades, 4 of spades, 5 of spades, 6 of diamonds

and so on.

Thank you,

John
 
Your English seems PERFECT...anyway, much superior to what we usually see here :rolleyes:

On your question:
I'm not sure what you mean:
same as 4 decks of 13 cards each, each labelled 1 to 13, all mixed into a 52 card deck?

Thank you Denis.

Yes, exactly that. Like we have 4 ace of spades, 4 two of spades ... 4 king of spades. No different suits.
 
Like we have 4 ace of spades, 4 two of spades ... 4 king of spades. No different suits.
You have thirteen ranks, 1 to king. But only four of each rank. You want to select six of them.

For a moment, suppose that we had at least six of each rank.
Then there are \(\displaystyle \dbinom{6+13-1}{6}=\dfrac{18!}{6!\cdot 12!}\) ways to do that.

But we have only four of each rank, thus we must remove the cases we don't want.
There are \(\displaystyle (13)(12)\) of those cases in which one rank is selected five times.
There are \(\displaystyle (13)\) of those cases in which one rank is selected six times.

So that means that we must remove those \(\displaystyle (13)^2\) impossible cases.
 
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