Bridge Hands: 5/6/2 Card Combination

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SUMMARY

The calculation of bridge hands consisting of 5 cards of one suit, 6 cards of another suit, and 2 cards of a third suit results in a total of 4,134,297,024 unique combinations. This is derived from the formula N = (4! / (4-3)!) * 13C5 * 13C6 * 13C2, which accounts for the permutations of the suits as well as the combinations of cards within each suit. The initial calculation of 172,262,376 only considers one specific combination of suits, necessitating the multiplication by the number of ways to choose 3 suits from 4 to generalize the result.

PREREQUISITES
  • Understanding of combinatorial mathematics, specifically combinations and permutations
  • Familiarity with the binomial coefficient notation (e.g., 13C5)
  • Basic knowledge of bridge card game rules and structure
  • Ability to perform factorial calculations
NEXT STEPS
  • Study advanced combinatorial techniques in card games
  • Learn about the application of permutations in probability theory
  • Explore the mathematical principles behind the binomial theorem
  • Investigate variations of bridge hands and their combinatorial implications
USEFUL FOR

Mathematicians, statisticians, bridge enthusiasts, and anyone interested in combinatorial game theory will benefit from this discussion.

Raerin
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A bridge hand consists of 13 cards. How many bridge hands include 5 cards of one suit, 6 cards of a second suit and 2 cards of a third suit?
 
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What if the question asked instead:

How many bridge hands include 5 cards of hearts, 6 cards of spades and 2 cards of diamonds?

Wold you be able answer that?
 
MarkFL said:
What if the question asked instead:

How many bridge hands include 5 cards of hearts, 6 cards of spades and 2 cards of diamonds?

Wold you be able answer that?

13C5 * 13C6 * 13C2 = 172,262,376

If my question is the same as this one then my textbook's answer key is wrong. The textbook says the answer is 4 xxx, xxx, xxx
 
Raerin said:
13C5 * 13C6 * 13C2?

If my question is the same as this one then my textbook's answer key is wrong.

Yes, good! :D That is correct, but this is for one specific combination of suits only.

Now you want to make it general. You want to multiply this by the number of ways to choose 3 suits from 4.
 
MarkFL said:
Yes, good! :D That is correct, but this is for one specific combination of suits only.

Now you want to make it general. You want to multiply this by the number of ways to choose 3 suits from 4.

I realized after I left that we need to find the permutations, not the combinations regarding the four suits, since order matters in this case because there are a different number of each suit. Hence, the number $N$ of the described bridge hands is:

$$N=\frac{4!}{(4-3)!}\cdot{13 \choose 5}\cdot{13 \choose 6}\cdot{13 \choose 2}=4134297024$$
 

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