Solving the Combinatronics Dilemma

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SUMMARY

The discussion addresses the combinatorial problem of selecting 5 cards from a standard 52-card deck while ensuring at least one card from each suit. Two methods were proposed: the first method calculates the combinations as 13C1 * 13C1 * 13C1 * 13C1 * 48, while the second method uses 13C1 * 13C1 * 13C1 * 13C2 * 24. The first method leads to double counting, as it treats different arrangements of the same set of cards as distinct selections, resulting in an incorrect total. The second method correctly accounts for the combinations without double counting.

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  • Understanding of combinatorial mathematics
  • Familiarity with binomial coefficients (e.g., "nCr" notation)
  • Basic knowledge of card games and suits
  • Ability to identify and correct counting errors in combinatorial problems
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  • Study advanced combinatorial techniques, including the principle of inclusion-exclusion
  • Learn about generating functions in combinatorics
  • Explore the concept of permutations versus combinations
  • Practice solving similar combinatorial problems using different methods
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Mathematicians, educators, students studying combinatorics, and anyone interested in solving complex counting problems.

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I solved a problem, by two different, and (apparently) viable solutions. But, each solution led me to a different answer:
The question asks me to find out the number of possible ways in which I can choose 5 cards from a standard pack of 52 cards, so that I choose at least a single card from each suite.

FIRST METHOD:
One card from each suite of 13 means a total of [13C1 * 13C1 * 13C1 * 13C1].
After that, I'll have 52-4=48 cards left from which I need to choose only one. That means 48C1.
The final result:
13C1 * 13C1 * 13C1 * 13C1 * 48

SECOND METHOD:
I'll be choosing one card from three suites, but exactly two cards from the fourth suite. So that means:
13C1 * 13C1 * 13C1 * 13C2
= 13C1 * 13C1 * 13C1 * 13C2 * (12/2)
And since I do this with each of the four suites (choose two from it only, while taking one from each of the others), I shall multiply this by 4, giving me:
13C1 * 13C1 * 13C1 * 13C2 * 24


And this is half of the answer arrived at by the first method...
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what went wrong?
 
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In the first method you double count. For example I could have picked my cards like this:

ace of hearts, ace of spades, ace of clubs, ace of diamonds

and then the king of heartsOR

king of hearts, ace of spades, ace of clubs, ace of diamonds

and then the ace of hearts

In the first method you count those as two distinct selections even though they're the same set of five cards
 
Oh...I see.
Thank you.
 

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