There was a statement made on a TV show that there were more unique shuffles of a deck of playing cards than there was sand on the beach. Using extreme limits, we can calculate whether this is true or not.
The Deck of Cards
First, can we calculate the number of unique shuffles of cards? That is a well-known process called combinations and permutations. It goes as follows:
The first card in every unique shuffle of a deck of cards must be one of 52 cards.
The second card will be one of 51 cards, since it cannot be the first card, which is already accounted for.
From this, we can determine that the first two cards in every unique shuffle must be one of `52*51` combinations. That is to say there are `52*51 = 2\ 652` possible combinations for the first two cards.
The third card can be one of 50 options. Each of these options comes with 2,652 possible combinations for the first two cards. This leaves us with `52*51*50 = 132,600` possible combinations for the first 3 cards.
We can continue this until we reach the last card. This is a mathematical operation called "factorial." The operational designator for the factorial operation is the exclamation point (!)
If we want to express the number of combinations of 10 items, we would write it as 10!
`10! = 10*9*8*7*6*5*4*3*2*1 = 3\ 628\ 800`
We can now calculate the number of unique shuffles of a deck of cards as follows:
`52! = 52*51*50*49* ... *5*4*3*2*1 = 8.065\ 817\ 5*10^67`
That is a BIG number! It is somewhat larger than...
`80\ 658\ 175\ 000\ 000\ 000\ 000\ 000\ 000\ 000\ 000\ 000\ 000\ 000\ 000\ 000\ 000\ 000\ 000\ 000\ 000\ 000\ 000`
If needed, we can include the EXACT number, but it will probably not be needed to answer the question.
Sand on the Beach
Can we calculate the number of grains of sand on the beach? Well, no. But we may be able to estimate it in a way that we know our estimate is at least as big as the number. If we can come up with a number that is absolutely LARGER than the number we are looking for, we might be able to answer the question.
First, lets make an extreme assumption. Lets assume the entire Earth is a beach.
Second, Lets assume the sand is super small sand. the size of a Hydrogen atom.
Third, lets assume the diameter of the Earth is larger than it is. Let's assume its diameter is uniform, but the same as the highest point on the planet. No, we will not use Mt. Everest because Mt. Everest is the highest point above sea level, but it is NOT the highest point from the center of the Earth. Since the Earth is an oblate spheroid, Mt. Chimborazo is more suitable for our purposes.
The absolute maximum volume of the Earth is:
`v=4/3 * pi *r^3`
`v=4/3 * pi *6\ 384.4^3\ km`
`v = 1\ 090\ 056\ 166\ 484\ km^3`
Grains of Sand on Earth
The diameter of a Hydrogen atom is 1.1 angstroms.
An angstrom is `1*10^-10` meters
If we round down, we can both simplify calculations and push the extreme limit out even further. So we will say the atom is 1 angstrom or 0.000 000 000 1 meters wide. This means 10 000 000 000 atoms will fit in a line one meter long. So about 10 000 000 0003 per cubic meter. That is 1 000 000 000 000 000 000 000 000 000 000 atoms per cubic meter.
That is 1.0`*`1030. So to calculate the absolute maximum number of sand particles on Earth, we simply multiply by the number of cubic meters of Earth. We get
`1*10^30*1\ 090\ 056\ 166\ 484 = 1.090\ 056\ 166\ 484*10^42`
While that is a big number, it is not as big as 52!. In fact, not only are there ABSOLUTELY more unique shuffles of cards than there is sand on the beach. We can calculate how many Earth-sized planets of sand it would take.
`(52!)/(1\ 090\ 056\ 166\ 484*10^42)=73\ 994\ 512\ 990\ 196\ 263\ 230\ 353\ 256`
In fact, it would only take a 37 card deck to consume the sand count of 8 Earth-sized planets.
No comments:
Post a Comment
Note: Only a member of this blog may post a comment.