MHB Bridge Hands: 5/6/2 Card Combination

  • Thread starter Thread starter Raerin
  • Start date Start date
  • Tags Tags
    Combination
AI Thread Summary
A bridge hand consists of 13 cards, and the discussion revolves around calculating the number of hands with specific card distributions: 5 cards of one suit, 6 of another, and 2 of a third. The initial calculation for a specific combination of suits yields 172,262,376 hands. However, to generalize this for any combination of suits, one must account for the permutations of choosing 3 suits from 4, leading to a revised total of 4,134,297,024 possible hands. The conversation highlights the importance of distinguishing between combinations and permutations in this context. The final formula incorporates both the suit selection and the distribution of cards among those suits.
Raerin
Messages
46
Reaction score
0
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?
 
Mathematics news on Phys.org
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:

$$N=\frac{4!}{(4-3)!}\cdot{13 \choose 5}\cdot{13 \choose 6}\cdot{13 \choose 2}=4134297024$$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Thread 'Imaginary Pythagorus'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...

Similar threads

Back
Top