MHB Bridge Hands: 5/6/2 Card Combination

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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.
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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:

$$N=\frac{4!}{(4-3)!}\cdot{13 \choose 5}\cdot{13 \choose 6}\cdot{13 \choose 2}=4134297024$$
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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