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

• Benzoate
So taking the natural log of 1 just gives you 0, which is why it is not necessary to include in the equation.
Benzoate

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?

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.

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

There are 8.0658 x 10^67 possible arrangements for a deck of 52 playing cards. This number is also known as 52 factorial or 52!, which is calculated by multiplying all the numbers from 52 down to 1.

2. How long would it take to go through all the possible arrangements for a deck of 52 playing cards?

If you were to go through one arrangement per second, it would take approximately 2.6 x 10^60 years to go through all possible arrangements for a deck of 52 playing cards. This is longer than the estimated age of the universe, which is around 1.4 x 10^10 years.

3. Is it possible to shuffle a deck of 52 playing cards into the same order twice?

The chances of shuffling a deck of 52 playing cards into the exact same order twice are extremely low. In fact, the probability of shuffling a deck into the same order twice is 1 in 8.0658 x 10^67, which is almost impossible to achieve.

4. How many unique combinations are there for a deck of 52 playing cards?

There are 52 unique combinations for a deck of 52 playing cards, as each card in the deck can only occupy one position at a time. However, there are 2.598 x 10^54 possible combinations, as the order of the cards can vary.

5. Can the number of possible arrangements for a deck of 52 playing cards be calculated using a formula?

Yes, the number of possible arrangements for a deck of 52 playing cards can be calculated using the formula n!, where n is the number of objects in the deck. In this case, n is 52, so the formula would be 52!. This is also known as the factorial formula.

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