# How many combinations for 3 cards from a deck of 52 cards if they have to be in ascending order?

If I wanted the different 3-card possible combinations from a total of 4 cards in ascending order, I would have:

$(1,2,3)$

$(1,2,4)$

$(1,3,4)$

$(2,3,4)$

EIDT:

How can I get all the possible, unique triplets in ascending order given an arbitrarily selected subset of cards from a standard deck of 52 cards?

I was misunderstanding the problem. Here's a bigger example, I'm given the cards:

$6, 1, 4, 3, 2, 4, 5$

• I can have more than one card with the same number, but each triplet cannot contain repeated numbers.
• Each triplet has to be in ascending order, but within the constraints of the original order.

In this example, the number 6 cannot appear in any triplet. The first 4 isn't going to appear in any triplet either, but the second 4 allows to form (1,4,5). The results are:

$(1,4,5)$

$(1,2,4)$

$(1,2,5)$

$(3,4,5)$

$(2,4,5)$

With these restrictions in place, it's probably not a combinatorics problem.

• Do you mean the standard deck of $52$ cards (which is the usual context) or do you mean cards numbered from $1$ to $52$ (which your example seems to suggest)? – Cameron Buie Jun 24 '13 at 4:31
• Counting is straightforward, it is $\binom{52}{3}$. As to an explicit list, "alphabetical order" sounds good. – André Nicolas Jun 24 '13 at 4:38
• Cameron, sorry about that, I meant 52 cards numbered from 1 to 52, will edit now. – sker Jun 24 '13 at 14:32
• Turns out, it is a standard deck of 52 cards. – sker Jun 24 '13 at 16:55

Hint: for any triplet there are $6$ orders and you accept $1$. Do you want the number of them or a list. The list will be too large for most purposes.
• @sker: For $n$ cards it is ${n \choose 3}$ because you once you select them, you can put them in exactly one order. – Ross Millikan Jun 24 '13 at 15:27
• With the edit, the deck is an existing sequence and you are looking for the number on increasing subsequences. There is no simple formula-$1$ to $52$ in order has ${52 \choose 3}$ subsequences, while if it starts in reverse order there are none. I think all you can do is count: start with the first element, look through the list for higher values to the right, then look for acceptable third values. Keep looping. – Ross Millikan Jun 24 '13 at 17:11