Does anyone know the formula for finding how many combos in a set of numbers?
367272
Suppose the number of combinations of k objects out of n objects without respect to order chosen is C(n, k). If we wanted to count order, we would multiply each combination by the number of ways it could be permuted. Thus P(k)*C(n, k) is the number of ways one could choose k people out of n people with regard to order. \nBut we can easily count this number another way. The number of possible choices for the first person is n, for the second person n-1, and so on until we reach n - k + 1. Thus, we get the equation P(k)*C(n, k) = P(n, k) = n*(n - 1)*...*(n - k + 1) = n!/(n - k)!. Solving for C(n, k), we get C(n, k) = n!/[P(k)(n - k)!] = n!/[k!(n - k)!]. \nNow you should be able to derive it whenever you get confused.
706198
5*4*3 = 60\n\nExample:\nNow let's suppose we have 10 letters and want to make groupings of 4 letters. It's harder to list all those permutations. To find the number of four-letter permutations that we can make from 10 letters without repeated letters (10_P_4), we'd like to have a formula because there are 5040 such permutations and we don't want to write them all out! \n\nFor four-letter permutations, there are 10 possibilities for the first letter, 9 for the second, 8 for the third, and 7 for the last letter. We can find the total number of different four-letter permutations by multiplying 10 x 9 x 8 x 7 = 5040. This is part of a factorial \nhttp://mathforum.org/dr.math/faq/faq.comb.perm.html
603689
add the numbers\n15\n19\n42\n19\nnext number equals 15..... and each has a 2 and a 7 in it\n\ntry 2742 sounds like a good answer for neopets lenny conundrum answer possibility but not fully sure..... lucidlyinsane
1157976
0000, 0001, 0002, 0003, 0004, 0005, 0006, 0007, 0008, 0009, 0010, 0011... etc, etc  untill u get to 9998, 9999\n\ntry it out, might take a long time but one combination will work :))
939247
No, but you can work it out on paper. I know because I did it once. You have five letters, so there should be 25 combinations, Of course, as two letters are the same that would make it 20. I think. Just give it a go. It's always more fun to work these tings out yourself.
469878
10 * 9 * 8 * 7 * 6 = 30240 combinations
754620
you can only get 6 - 36.you can only get 6 one way, and 36 one way.  you get 7 six ways, and 35 six ways.  figure out all the ways to get each number, including 19, and add them all up.  the odds of getting 19 is the number of ways to get 19 over the total number of outcomes
741355
There are a few ways to find this, but the answer depends on whether we can repeat numbers in the combination and if the order matters.  \n\nIf we cannot repeat numbers and order matters:\n# combinations = 9*8*7*6*5*4 = 60480\nThere are 9 choices for the first digit, 8 for the second (because one is already taken), and so on.  We multiply the choices for each digit to find the total number of choices.\n\nIf we can repeat numbers and order matters:\n# combinations = 9^6 = 531441\nThere are 9 choices for each of the six digits.\n\nIf we cannot repeat numbers and order does not matter:\n# combinations = 9!/(6!*3!) = 84
791300
Use the FOIL method; First, Outer, Inner, Last.\n\nFirst, n * n\n\nOuter, n * 5\n\nInner, n * 7\n\nLast 7 * 5.\n\nThen add them all together.\n\n22 NOV 06, 0420 hrs, GMT.
293919
The answer is 56.\nAll the letters in the word PROLEMS are distinct. So we have 8 different letters available with us. So the answer is\n8_C_3 i.e. 8!/[(8-3)!*3!]=8*7*6/(3*2)=8*7=56\n\nThe formula for combinations, choosing 'r' from 'n' is\n          \nn_C_k = n! /[ k!(n - k)!]\n\nHere n! stands for 'n factorial'\nFactorial for a postivie integer 'n' is the product of all integers\nform 1 to n.
804992
Let us call the first 2 "2 sub 1", the second "2 sub 2" ... same thing for the 6's.\n\nLet us say that "2 sub 1" is distinct from "2 sub 2", and so on.\n\nWe have 8 choices for the first number, seven choices for the second number, etc. This means that there are 8*7*6*5*4*3*2*1 choices, or 8!. \n\nNow if you look at any number, take all of the 2's out. You see that you can rearrange the 2's in 3! different ways (including the original arrangement) using the procedure in the paragraph above. If we take out the 2's rearrange them, and put them back in we have created the a new arrangement. So for each arrangement of numbers, we have counted it 3! times. So we divide by 3!. The same thing for the sixes, we divide by 2!.\n\nThe result it (8!)/((3!)(2!)) = 3360 choices.
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
182985
No. of alphabets = 22  [{a, b, .. z) - {i, o, q, z)]\n==> n = 22\nNo. of letters taken at a time = 3 \n==> r = 3\n\nNo. of combinations of "n" different things taking "r" at a time is given by n_C_r or C(n,r) \n= (n)(n-1)..r times/(r) * (r-1) * ...(1)\n\nTherefore, the number of combinations that can be formed \n= C(22,3)\n= 22 * 21 * 20/3 * 2 * 1\n= 1,540\n\nIf you are looking for the number of words that can be formed using the letters,.. then you have to look out for permutations..\n[No. of permutations of "n" different things taking "r" at a time given by n_P_r or P(n,r) = (n) * (n -1) * ... r times.\n\nThe number of words that can be formed \n= P(22,3)\n= 22 * 21 * 20\n= 9,240\n\nIf the same letter can be used then the number of words that can be formed is given by (n)^r\nThe number words that can be formed \n= 22^3\n= 22 * 22 * 22\n= 10,648\n\n\nhttp://www.futureaccountant.com/probability/\n\nhttp://www.schoolingkids.com/
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:(
110013
Well, all numbers would have to be represented equally, right?  If there are 27 combination I assume that 1, 1, 1 is included in these combintations, not just every combination that includes all 3 numbers...  However, if all three numbers are represented equally, each number should appear 27 times.  So the answer should be 27*9, or 243.
102578
http://home.howstuffworks.com/combination-lock.htm
16141
There are 6 combinations\n3,5,7\n3,7,5\n5,3,7\n5,7,3\n7,5,3\n7,3,5