Combinations and Permutations of Cards

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Discussion Overview

The discussion revolves around the problem of dividing a deck of 52 cards into piles of 3, where each pile must contain a positive number of cards. Participants explore the mathematical approaches to calculate the number of ways to achieve this division.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant asks how many ways a deck of 52 cards can be divided into piles of 3, with each pile containing any number of cards.
  • Another participant questions whether a pile can contain 0 cards and suggests a method for calculating the combinations based on the value of k, the number of cards in the first pile.
  • A participant clarifies that 0 is not allowed in any pile, leading to a different approach to the problem.
  • Further clarification is requested regarding the summation process for k ranging from 0 to 52.
  • Another participant explains that if 0 is not allowed, the number of ways to sort the remaining cards increases as the number of cards in the first pile decreases, leading to a summation of integers from 1 to 50.

Areas of Agreement / Disagreement

Participants agree that 0 is not allowed in any pile, but there is no consensus on the overall method for calculating the combinations, as different approaches are being discussed.

Contextual Notes

The discussion includes assumptions about the treatment of piles and the implications of allowing or disallowing 0 cards in a pile, which may affect the calculations presented.

Baron
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Hey guys,
I have a problem relating to combinations and permutations.In how many ways can I divide a deck of 52 cards into piles of 3 with each pile containing any number?
for example 50,1,1 or 45,6,1
Thanks in advance
 
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Is 0 allowed for one of the piles? If you look at in two steps it is fairly straightforward.
Step 1, choose a number (k) for the first pile. There are now 52-k left for the other two piles. Assuming 0 is allowed, there are 52-k ways to split up these cards. Now just add them up for k ranging from 0 to 52. It will sum to 52x53/2.

If 0 is not allowed, the procedure is the same, just omit the cases where any pile has 0.
 
Thanks,0 is not allowed
 
Can you explain this."Now just add them up for k ranging from 0 to 52".I'm a little lost.(not a math expert)
 
Since 0 is not allowed. Put 50 in the first pile, then there is exactly 1 way of sorting the other two piles. Put 49 in the first pile and there are 2 ways of sorting the other 2 piles, ..., put 1 in the first pile and there are 50 ways of sorting into the other two piles. Net result 1+2+3+...+50 = (50x51)/2.
 

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