how to find the number of combination of one pair when 5 cards are given out?
539344
I assume the question you are asking is how many possible five card hands contain at least one pair.\n\nThere are 13 face cards from which the pair can come.  Once you have chosen from among the 13, there are (4C2) ways to choose the two suits.  There are 52-4=48 remaining cards, and you need to choose 3, so there are (48C3) possibilities for the other three.\n\nThe answer is 13 x (4C2) x (48C3) possible five card hands containing at least one pair.
672664
first you figure out the number of possible outcomes you can have. which would be 6x6 = 36. \nThen you figure out how many combinations you can make to make it greater than 7:\n5 and 3\n5 and 4\n5 and 5\n4 and 4\n4 and 5\n3 and 5\nwhich is 6.\n6 over 36 is one sixth. therefore your answer is 1/6.
970878
OOOOOOOOOh i LUV LUVLUVLUV challenge 24!!!! ok her are some ideas:\n7,2,2,1=(7-1)(2+2)\n6,8,2,1=(8/4)(6*1)              ***All taken from Challenge 24 cards\n3,3,6,2=(3+3)(6-2)                     (single digit)\n9,2,5,1=(9-1)(5-2)\n~* ah what the heck i dont have time to finish this. I'llget back to you later, soon. and for future reference, DO buy challenge 24 cardsand so you can use them often. the only annoying thing is, it doesnt come w/anser key u have to figure it out yourself:(
939050
Assuming order matters on the first 5 numbers, there are 39 x 39 x 39 x 39 x 39 x 19 combinations of draws.  Only one matches with your selection making the odds\n1 in 1,714,259,781
375114
To have four sixes in a hand of five means there are only 48 cards left to choose from as the last card. So 48.
236101
permutation?\n\ni'm not sure what you mean by "see how many it is."
93645
d=5, g=1\n\nFirst I found all the combinations of 3 numbers (no repeats and no fours) that added to 13 (17-4) for b+c+d. Then I found all the combinations of 3 numbers (no repeats and no fours that added to 17 for g+h+i. Then I figured out which of these pairs of numbers had digits that didn't overlap, and had a pair of digits (d & g) that added to 17 minus the digits that, between them, were still missing (e & f).\n\nAnd, I'm a gal. :)
527818
The Fundamental Counting Principle gives us 12 combinations in this case, so if the first 12 are wearing different things, then the 13th must be wearing one of the other 12's combos. The answer to your second question is 25.
321839
farrell the first part of your solution is wrong.  There are 10 places to put the vowels, not 8, they can be at the beginning or the end. I believe it is 9! * (10!/5!) = 9! * 10 * 9 * 8 * 7 * 6\n\ntsniperx might be right but the question is ambiguous about this.  It says how many arrangements of the cards (not sequences of letters) are there which to me means that 0'0 and 00' are two different arrangements because the cards are in different order...but like I said it isn't obvious either way.  \n\nIF tsniperx is correct cornelius is still wrong, it would be 9!*10!/(5!*2!*2!*3!)  In other words, cornelius' answer is double the correct answer.  But I still think my original answer is correct.
761643
How many at a time?  You didn't say.\n\nThe number is an expression called  nCr, which is \nn!/((n-r)!r!), but it's pretty simple to think of it this way:\n\nI'll suppose you want them 3 at a time.  Then, for the first one, with 6 letters to choose from, write 6*5*4 on top (three factors because we want 3 letters) and 1*2*3 on the bottom\n6C3 = 6*5*4/(1*2*3)\n        = 20\n\nIf there had been 10 letters and we wanted to know the number of combinations 4 at a time, we would put four factors on top and bottom:\n10C4 = 10*9*8*7/(1*2*3*4)\n          = 210
106376
Also, if you agree that the formula for the number of combinations of n objects taken r at a time is\n\nn!/(r!(n-r)!)\n\nand that there is only one way to get a combination of n objects taking all n at a time, you'd get\n\nn!/(n!(n-n)!)  or n!/(n!0!), or canceling, 1/0! which must equal 1.\n\nSo 0! must equal 1.  More of a demonstration than a proof but it's easy to understand.
339616
There are 12 face cards.\nThere are 26 red cards.\nThere are 6 face cards that are red.\n12+26-6 = 32\n32/52\n= 8/13
1071317
There are exactly 343 combinations.  I can list them all, but others have already cluttered this thread quite well on their own.  ^.^\n\nYou can think of the solution set as a series of tiered combinations.  These would start with those sets beginning with a $50 bill, which can contain up to 2 x $50,  2 x $20, 5 x $10, etc. which I believe someone has started already in this thread.  Or group them as pairs, triplets, etc. with the equation 50j+20k+10l+5m+1n=100, which may be a bit easier for calculations.  Or a convergent series, where A=(ways to form $5), B=(ways to form $10, including A-sets), etc.  With any method, find the different combinations for each set, then add them all up.\n\nDoing this is still tedious and prone to mistakes, and thus probably not much of an improvement over manually listing all of the combinations one by one.  It can also quickly become an insane exercise for larger values (for example, add the $2 bill, or find change for $200, or even worse).  So instead of trying to think this problem through mathematically, I found it much easier to think it through as a programming problem.  It lends itself very readily to a simple recursive subroutine using the following pseudocode:\n\nmake an empty stack\nset total combinations to 0\ncall CheckCurrency\nprint total combinations found\n\nrecursive subroutine CheckCurrency\n- if total currency in stack = 100\n- - add 1 to the total combinations\n- - (optional: display stack contents)\n- - end recursion\n- if total currency in stack > 100\n- - end recursion\n- if total currency in stack < 100, then\n- - for each valid denomination (50,20,10,5,1)\n- - - if the denomination < = smallest denomination in the stack\n- - - - push it to the stack\n- - - - call CheckCurrency\n- - - - pop it from the stack\nend of subroutine\n\nThis required only a few minutes to design, another few minutes to code, another few minutes to debug, less than 2 seconds to run, and has the wonderful bonus of being able to display every single combination if desired.  It would also be trivial to modify it so that, for example, you can find all combinations if you also include the $2 bill.  (The answer is:  4562 combinations.)  Or to find all combinations totalling $200.  (The answer is: 2930 combinations.)\n\nThe line that checks if the denomination is less than or equal to the smallest denomination currently in the stack is present to avoid a common mistake:  We want the total combinations, not total permutations.  Otherwise, the stack "50 20 20 10" would be counted differently than the stacks "50 20 10 20" and "50 10 20 20", and even "10 20 20 50".  The total permutations is much, much greater, especially if you consider each bill as unique.  Five $20 dollar bills alone have 5!=120 permutations because, after all, they are all completely different $20 bills.  Imagine the number of permutations of 100 x $1 bills, and everything in between.  o.O
838732
000000 to 999999 --> 1 million combinations\n\n10 choices for the first digit\n10 choices for the second digit\n10 choices for the third digit\n10 choices for the fourth digit\n10 choices for the fifth digit\n10 choices for the sixth digit\n\n10 x 10 x 10 x 10 x 10 x 10 = 1 million\n\nHowever, if you disallow 0 as a starting digit then you have the following:\n\n*9* choices for the first digit (1-9)\n10 choices for the second digit (0-9)\n10 choices for the third digit (0-9)\n10 choices for the fourth digit (0-9)\n10 choices for the fifth digit (0-9)\n10 choices for the sixth digit (0-9)\n\n9 x 10 x 10 x 10 x 10 x 10 = 900,000
724915
simple.  if you want to find the permutations of something of a set of numbers (or letters or objects), as long as you want to find just how many combinations there would be, you use factorial (!).  8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1...and that's your answer.
922466
54 - if you count the jokers\n+12\n+190\n---------\n256\n----------\n\nor     52\n     + 13 - counting the number 12 as two seperate numbers\n     +190\n    ---------\n        2 5 5\n   ------------\n\nor 54+13+190= 257\n\nor Cards 52 + numbers on clock(counting each single digit) 15+ 190 = 257 with jokers in a pack of cards = 259\n\nso the answer is either 257, 259, 254, 256, 86 or 88\n\noooo! My head hurts!\n\n=kizzy=
938617
You can read off the outcomes from Pascal's Triangle. For 5 tosses consult the fifth row\n\n........1  1\n.......1  2  1\n......1  3  3  1\n.....1  4  6  4  1\n....1 5 10 10 5 1\n\nThere is \n1 chance in 32 of 5 heads and 0 tails\n5 chances in 32 of 4 heads and 1 tail\n10 chances in 32 of 3 heads and 2 tails\n10 chances in 32 of 2 heads and 3 tails\n5 chances in 32 of 1 head and 4 tails\n1 chance in 32 of 0 heads and 5 tails