Solving the Combinatronics Dilemma

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Two different methods were proposed to solve the problem of selecting 5 cards from a standard 52-card deck, ensuring at least one card from each suit. The first method calculated the combinations by choosing one card from each of the four suits and then selecting an additional card from the remaining 48, leading to a potential overcounting of identical sets. The second method involved selecting three cards from three suits and two from one suit, which correctly accounted for the combinations without double counting. The discrepancy between the two methods highlighted the issue of overcounting in the first approach. This discussion emphasizes the importance of careful combinatorial reasoning in solving such problems.
modulus
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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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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