How many possible arrangements are there for a deck of 52 playing cards?

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SUMMARY

The total number of possible arrangements for a standard deck of 52 playing cards is calculated as 52 factorial (52!). This represents the vast number of permutations available when shuffling a sorted deck. The entropy created through this process can be expressed using the formula ΔS = k(ln(52!) - ln(1)), where k is the Boltzmann constant. The discussion also references the source of the omega equation from Daniel V. Schroeder's "Thermal Physics" textbook, emphasizing the significance of understanding entropy in relation to thermal energy arrangements.

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  • Understanding of factorial notation and permutations
  • Familiarity with the concept of entropy in thermodynamics
  • Knowledge of Boltzmann's constant (k) and its application
  • Access to Daniel V. Schroeder's "Thermal Physics" for reference
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This discussion is beneficial for students in physics, mathematics, and computer science, particularly those interested in thermodynamics, statistical mechanics, and combinatorial analysis.

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Homework Statement


How many possible arrangements are there for a deck of 52 playing cards?(For simplicity, consider only the order of the cards , not whether they are turned upside down, etc.) suppose you start with a sorted deck and shuffle it repeatedly , so that all possible arrangements becomes accessible? how much entropy have you created in the process? express your answer both as a pure number(neglecting the factor k) and the SI units.Is the entropy significant compared to the entropy asociated with arranging thermal energy among the molecules in the cards?

Homework Equations



S=k*ln(omega), k is neglected in this problem.

omega=(q+N)!/((q)!(N)!)

The Attempt at a Solution



a)How many possible arrangements are there for a deck of 52 playing cards?

the number of arrangements is just N factorial or in my case 52!

b)suppose you start with a sorted deck and shuffle it repeatedly , so that all possible arrangements becomes accessible how much entropy have you created in the process?

so would I just calculate the total number of omega 's: In other words, would I calculate all the posible q's? for instance , omega(q=0)=(0+52)/((0!)(52!)+omega(q=1)=(1+52)/((1!)(52!))+...+omega(q=51)=(51+52)!/((51!)(52!))+omega(q=52)=(52+52)!/((52!)(52!)) and then proceed to take the natural log of all the total sums of the omega's to calculate my entropy?
 
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I'm not sure where your equation for omega comes from, but omega is supposed to just be the number of possible arrangements. So your increase in entropy is just:

\Delta S = k \left( \ln 52! - \ln 1 \right)
 
genneth said:
I'm not sure where your equation for omega comes from, but omega is supposed to just be the number of possible arrangements. So your increase in entropy is just:

\Delta S = k \left( \ln 52! - \ln 1 \right)

Is my calculation for the total number of arrangements correct? I do not understand how you obtained ln 1.

i obtained my equation for omega on p. 63 of Daniel's V. Schroeder Thermal physics textbook.
 
Last edited:
The number of arrangements is correct. The 1 comes from the fact that there is only one way to arrange the deck in a sorted manner.
 

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