What is the number of possible orderings of a deck of 52 playing cards?
So there are 52 cards, how many possible sequences are there of 52 using each card only once?
Answers:
That number equals 52! (fifty-two factorial). It may seem surprising that this number (80,658,175,170,943,878,571,66. is so big. Comparing that number to some other large numbers, it is greater than the square of Avogadro's number, 6.022 × 1023.
i think it is 52^52 power
use a scientific calculator.
2,704
well i gave up pressin buttons at x9.and at x10 my calc went haywire..but its 2.20am here, so give me a break, and at least i tried.
good luck.
52!
Or 52 factorial
8.0658175170943878571660636856. x 10^67
That's 52! (52 factorial)
You've got 52 choices for the first card, times 51 choices for the second card, times 50 choices for the third card .
8.066 x 10 to the power 67
Or 8066 followed by 64 zeros (roughly)
52! or 52*51*50*49*48*47.7*6*5*4*.
80,658,175,170,943,900,000,000.
Why am I also thinking it's 52 times 52 - ? (52 squared) My other answer is if you were to drop them all on the floor you would only be limited to your brain's imaginings the number of orderings you could have(because you could order them all with your eyes only), but I think there is a mathematical answer which would of course supersede my foolish remarks. I think it's a darned shame there is a practical, down-to-earth, sane and definite answer to this one. It would be so much more fun if you introduced some additional dimension, so the answer would then become an infinite number of orderings. - C. (Quantum mechanics, anyone?)
52 x 51 x .. x 1 ways.
If you have n different numbers to order there are n x (n-1) x .. x 1 ways to do so. In math they write this as n!
If you have y numbers the same, divide n! by y!
For example there are 52 ppl but among these 5 are quintets, there are (52 x 51 x ..x 1) / (5 x 4 x..x 1) ways to order them.
If there are 52 ppl, among these there are 7 groups of quintets, there are (52 x 51 x..x 1) / [7 x (5 x 4 x ..x 1)] ways.
So if you want to order 52 cards where suit doesnt matter, there are 52! / (13 x 4!)
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Answers:
That number equals 52! (fifty-two factorial). It may seem surprising that this number (80,658,175,170,943,878,571,66. is so big. Comparing that number to some other large numbers, it is greater than the square of Avogadro's number, 6.022 × 1023.
i think it is 52^52 power
use a scientific calculator.
2,704
well i gave up pressin buttons at x9.and at x10 my calc went haywire..but its 2.20am here, so give me a break, and at least i tried.
good luck.
52!
Or 52 factorial
8.0658175170943878571660636856. x 10^67
That's 52! (52 factorial)
You've got 52 choices for the first card, times 51 choices for the second card, times 50 choices for the third card .
8.066 x 10 to the power 67
Or 8066 followed by 64 zeros (roughly)
52! or 52*51*50*49*48*47.7*6*5*4*.
80,658,175,170,943,900,000,000.
Why am I also thinking it's 52 times 52 - ? (52 squared) My other answer is if you were to drop them all on the floor you would only be limited to your brain's imaginings the number of orderings you could have(because you could order them all with your eyes only), but I think there is a mathematical answer which would of course supersede my foolish remarks. I think it's a darned shame there is a practical, down-to-earth, sane and definite answer to this one. It would be so much more fun if you introduced some additional dimension, so the answer would then become an infinite number of orderings. - C. (Quantum mechanics, anyone?)
52 x 51 x .. x 1 ways.
If you have n different numbers to order there are n x (n-1) x .. x 1 ways to do so. In math they write this as n!
If you have y numbers the same, divide n! by y!
For example there are 52 ppl but among these 5 are quintets, there are (52 x 51 x ..x 1) / (5 x 4 x..x 1) ways to order them.
If there are 52 ppl, among these there are 7 groups of quintets, there are (52 x 51 x..x 1) / [7 x (5 x 4 x ..x 1)] ways.
So if you want to order 52 cards where suit doesnt matter, there are 52! / (13 x 4!)
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