My Math Forum Card Combinations

 September 15th, 2012, 08:15 AM #1 Newbie   Joined: Sep 2012 Posts: 2 Thanks: 0 Card Combinations I have a problem that needs solving: From a pack of 52 cards, how do you select three cards that have these specific properties. 1. Two of these cards must be of the same rank. 2. The third of these cards must be of a rank that is two gaps higher or lower than the other two cards. 3. All three cards must be of a different suit. 4. The result must be expressed as a combination (not a permutation). So the combination 6spades 6clubs 9hearts fits in this category. As does 6hearts 6diamonds 3clubs also fit in this category. Instead of just producing an answer I would really appreciate if someone could explain the reasoning behind the result.
September 15th, 2012, 09:42 AM   #2
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Re: Card Combinations

Quote:
 Originally Posted by nuerofuzz 2. The third of these cards must be of a rank that is two gaps higher or lower than the other two cards. So the combination 6spades 6clubs 9hearts fits in this category. As does 6hearts 6diamonds 3clubs also fit in this category.
Why do your examples show 3 gaps?

KingHearts KingSpades ... is there a HIGHER gap card?

AceHearts AceSpades ... is there a LOWER gap card?

 September 15th, 2012, 12:24 PM #3 Newbie   Joined: Sep 2012 Posts: 2 Thanks: 0 Re: Card Combinations Just to claritify, 663 represents a two card gap. The gap would comprised of 5 and 4. As would be the case with 669, the gap being 7 and 8. And because the 13 card rankings can be counted cyclically then one could have AA4 or AAJ. Both combinations would represent a pair and a two card gap.
September 15th, 2012, 12:49 PM   #4
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Re: Card Combinations

Hello, nuerofuzz!

Quote:
 From a pack of 52 cards, how do you select three cards that have these specific properties. 1. Two of these cards must be of the same rank. 2. The third of these cards must be of a rank that is two gaps higher or lower than the other two cards. 3. All three cards must be of a different suit. 4. The result must be expressed as a combination (not a permutation). So the combination $6\spadesuit,\,6\diamond,\,9\clubsuit$ fits in this category. As does $6\diamond,\,6\clubsuit,\,3\spadesuit$ also fit in this category. And the ranks are cyclic.

$\text{The first card can be any card: }\,_{52}C_1 \,=\,52\text{ choices.}$

$\text{The second must be the same rank as the first: }\,_3C_1\text{ choices.}$

$\text{The third must be 3 ranks higher or 3 ranks lower than the first card}
\;\;\text{and be of a different suit: }\,_4C_1\text{ choices.}$

$\text{Therefore, there are: }\:\left(_{52}C_1\right)\left(_3C_1\right)\left(_4 C_1\right) \;=\; 624\text{ ways.}$

 September 15th, 2012, 08:50 PM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: Card Combinations 6C 6D ; 3H or 3S or 9H or 9S = 4 ways ; similarly for 5 more: 6C 6H 6C 6S 6D 6H 6D 6S 6H 6S So that's 6 * 4 = 24 With 13 ranks: 13 * 24 = 312 How do you get 624, Soroban?
September 15th, 2012, 10:06 PM   #6
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Re: Card Combinations

Hello, Denis!

You're absolutely right!

Quote:
 Originally Posted by I $\text{The first card can be any card: }\,_{52}C_1 \,=\,52\text{ choices.}$ $\text{The second must be the same rank as the first: }\,_3C_1\text{ choices.}$

I imparted an order to the first two cards . . . *blush*

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