In order of number and suit, just as if you openned the deck for the first time. Also assuming that you aren't really trying to shuffle them into order. Based on the odds of this being a happenstance.
128914
Shuffling is pretty complex as you probably mean a riffle shuffle or something like that.  But let's assume you're shuffling by spreading out all the cards on the table, and randomly picking them.\n\nAssuming you want the cards in a specific order, then for each card, you'd have to randomly select the "right" card.  For the first card, you have 52 to choose from, and a 1/52 chance of getting it right.  For the second card, you now have 51 to choose from and thus a 1/51 chance of getting it right...on down to the last card.\n\nSo your odds are:\n\n1/52 * 1/51 * 1/50 * 1/49 * ... * 1/3 * 1/2 * 1/1\n\nor 1/52! or about 1/8*10^69.  To put that in perspective, lets say all 6 billion people on earth are shuffling one trillion times a second (man is that fast shuffling!) for a billion years, you're still not even in the ball park of being close (1/10^30) of having one person get it right.
43424
Combinations are basically what you do when order doesn't matter.For example, electing any 12 students to a group of 3. It would be 12 C 3.\n\n\nPermutations are what u do when order DOES mattter. For example, when your arranging 6 different books. The answer to it would be 6 x 5 x 4 x 3 x 2 x 1 = 720 different ways.\n\nFactorials are basically. The number you have multiplied by everything less than that number.For instance, if you have 3. The factorial notation would be 3 x 2 x 1 = 6
643710
#4, #2, #4, #4, #2, #4 #1...repeat......
266527
I hope you mean 52 cards.\n\nAnyway, to figure out this problem, you would need to find the number of combinations of 5 taken from 52. The equation for that is:\n52!/(52-5)!5!\nThis is equal to 52!/47!5!\nWhich is equal to: (48x49x50x51x52)/(2x3x4x5)\nThere are a total of 2,598,960 poker hands. \n\nIt's best to find the number of ways to get three kings, then two queens, then to multiply the two. \nThere are four possible kings and combinations of three. \n4!/(4-3)!3! This equals 4. \n\n4!/(4-2)!2! This equals 6. \n6x4=24\nThere are 24 possibilities of three kings and two queens. \n\n(c) is a little harder to explain. For every card Ace thru King there are 12 other cards it could make a full house with. The king can make 24 with a queen, but it can also make 24 with ace, two, three, four etc. To find the total number one must multiply 24 by 12. This equals 288. Now we must multiply that number by the total number of different kinds of cards in the deck. That would be 13. 288x13 = 3,744.\n\nThere are 3,744 ways to make a full house. \n\nTo find the probability, one must use the answer from question (a) and put the answer to question (c) over it. \n\n3744/2598960 =\n\n6/4165.\n\nOr an approximately 0.144 % chance of being dealt a full house.
563748
They should all be aprox. the same.  If they are not, you probably need to straighten something out.
661365
In all these problems you are going to pick a set of objects or perform a set of actions.  \n\nIf the ordering of the outcomes makes a difference, then it's a permutation. If the ordering doesn't make a difference, then it's a combination.\n\nExample 1: The 10 runners run a race. They give medals for the first 3 places. How many different outcomes are there?\n\nThe order of finish makes a difference, since it determines who gets which medal, so it's a permutation problem.\n\nAnswer: 10x9x8 = 720\n\nExample 2: Three students are chosen from a class of 10 to clean after a food fight. How many ways can the group be chosen?\n\nIt doesn't matter who gets picked first, they all have to clean. Ordering doesn't make a difference, so it's a combination problem.\n\nAnswer: 10C3 = 10!/3!7! = 10x9x8/3x2x1 = 120.
46349
Wrong!  It's not 1st and then 10th if there's ten teams....  It goes, from 1st pick to 9th pick, then the 9th team gets first pick in the second round and the 1st team gets the last pick in that round.  So, the 9th team would get the 10th pick and the 1st team would get the 18th.  \n\nIt's done that way so it's fair.  Think about it - if you were the last team to pick, would it be fair if you had to pick last EVERY time, and the 1st team got to pick FIRST?  So no, it switches every round.
237336
There are 20! possible orderings for picking the numbers from the hat. Only one has them in order, so 1/20!.\n\nAnother way to look at it is that when there are 20 numbers in the hat, your chances are 1/20 of picking the 1. Then they are 1/19 of picking the 2, then 1/18 of picking the 3 etc.\n\nSo the chances are (1/20)(1/19)(1/18)...(1/2)(1/1)\n\nIf you threw the numbers back in each time, the probablity of pulling them out in order would be 1/(20^20).
148839
I think it is order(N squared), so not very good. It's a variant of bubble sort, which you can google. Plus you never know when you're done!
579128
you can only fold it in half once.  after that, you're folding it in quarters, eigths, etc
543575
i always placed them in order i got them...from the top down...some placed them in order of importance to them..of course if some one won state that would be the first!! but mine was a M
587889
1/(52!)\nThere's only one correct order.\nThe number of ways to arrange 52 objects is 52!
161258
The best way of covering all 15 numbers evenly is simply to start by selecting 1,2,3,4,5,6 then 7,8,9,10,11,12 then 13,14,15,1,2,3 etc. That way, you're sure to hit each number an equal amount of times in the long run. If you're saying "but these combinations are totally unlikely to be selected!" then yes, you are right. But they are precisely just as likely as any other of the combinations. So, in other words, don't play the lottery.
620429
Most of them got it right.\n\nThis is another way to think about the result.\n\nFirst player = T or H doesn't matter\n\nSecond Player = Must be that same as the First Player\n                        =  only 2 possibility (H or T)\n                        = 1/2\nThird Player = Must be the same as First Player\n                    = again, only 2 possibility ( H or T )\n                    = 1/2\n\nIn order for the Second Player and Third Player to have the same result as First Plaer = 1/2 x 1/2\n                                 = 1/4.
156850
Mine are alphabetized, then each band is put in chronological order for the studio albums. If two bands share a band member, or if a band splits up and one of them does a solo album, I will put them together. \n\nWith singles, I organize them between the studio albums, just after the studio album they were released off of. If I know the order in which they were released I will put them in chronological order. So it may be album, then 4 singles, 2nd album, 3 singles, 3rd album... and so on. Live music goes last in my organization.\n\nIt sounds like a lot of work but if you have a music encyclopedia to look up the band member associations, it can be a really cool project to accomplish. And you can learn a lot!
11557
There is no rule like that.  You can check as many times as you want except if the same exact position is repeated three times.  Then it's a draw and the game ends as a tie.
596795
Think about your order of operations. If you were evaluting the function for some number, what would you do first? That function is your *inside* function.