how do i find the sum of all the combinations of 1, 3, 5?  I understand there are 27 combinations, but sum???
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.
464956
use each number once and only once...\n\n2*3+4*5=26   \n\nThis is too easy.  Give us a hard one!\n...come on...?
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
109803
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
855230
Hope this is helpful...\n\nBasically you can get the number of outcomes by looking at the corresponding coefficient in the pascal triangle.\n\n#1) 10 (row 5, 4th number) \n#2) 1+6+15+20=42 (add up 1st 4 numbers in row 6)\n#3) 10 (row 10, 2nd number)\n#4) 1+8+28+56+70=153 (add up 1st 5 numbers in row 8)
982580
Use binomial coefficients to solve this.\n\n| n |           n!\n|    |   =  ----------\n| k |        (n-k)!k!\n\nn is the number of possibilities and k is the number of unordered outcomes.\n\nCombination of six numbers from ten possibilities.\n\n| 10 |       10!          3628800         \n|      | = ------------ = --------------- = 210\n|   6 |    (10-6)!6!     (24)(720)\n\nCombination of four numbers from ten possibilities.\n\n| 10 |\n|      |\n|   4 |\n\nYou should be able to figure out the second one on your own.\nHere's a link on factorials, just in case you don't know what they are.\n\nhttp://mathworld.wolfram.com/Factorial.html
242452
This is a typical combination problem.\n\nformula for combination is n!/r!(n-r)!\n\n= 9! / 5!(9-5)!\n= 9! / 5!4!\n= 126 (ANS.)\n\nfactorial ex. 5! = 5 x 4 x 3 x 2 x 1\n\nDONT FORGET TO RATE THIS AS BEST ANSWER
837543
If I have interpreted you correctly, each of the 49 packages must contain either BBB, BBS, BSS, or SSS.\nThen there are:\n22,100 combinations.\n\nThis is the same as the number of solutions of the equation:\nx1 + x2 + x3 + x4 = 49\nin non-negative integers.\n(i.e., x1 is the number of BBB packages, x2 is the number of BBS, x3 the number of BSS, and x4 the number of SSS)\n\nThis is C(49+4-1, 4-1) = C(52,3) = 52*51*50/(3*2*1) = 22,100.\n\n*************************\nADDITIONAL DETAILS:\nOk, I saw that you reposted this question.  However, you are still missing something or have the numbers or the question incorrect.\n\n285 = 3*5*19 .  This number is not a result formable by the combination function: C(a,b) = a!/[(a-b)!b!]\n\nThe closest would be 286 = C(13,3).  This would be the number of ways that 3 items could be chosen from 13 items.\n\nAlso, as you have it written on the other post, my answer above is still correct.  There are 22,100 ways that 4 different types of items could be distributed into 49 packages.  This assumes that the packages are indistinguishable from each other.  If they weren't, i.e., if BBB in package #1 and BBS in package #2 is different from BBS in #1 and BBB in #2, then the number of combinations is MUCH larger.  (It would be 4^49 ~ 3.2 * 10^29)
361589
Sum of all numbers 1 to 100:\nS = n[2a1 + (n - 1)d]/2\n= 100[2(1) + (100 - 1)]/2\n= 100(101)/2\n= 5050\n\nSum of all numbers 3 to 99, multiples of 3:\nS = n[2a1 + (n - 1)d]/2\n= 33[2(3) + (33 - 1)3]/2\n= 33(6 + 96)/2\n= 1683\n\nSum of all numbers 5 to 100, multiples of 5:\nS = n[2a1 + (n - 1)d]/2\n= 20[2(5) + (20 - 1)5]/2\n= 20(10 + 95)/2\n= 1050\n\nSum of all numbers 15 to 90, multiples of 15:\nS = n[2a1 + (n - 1)d]/2\n= 6[2(15) + (6 - 1)15]/2\n= 6(30 + 75)/2\n= 315\n\nSum of integers from 1 to 100 which are not divisible by 3 and 5:\nS = sum(1-100) - sum(3-99) - sum(5-100) + sum(15-90)\n= 5050 - 1683 - 1050 + 315\n= 2632\n\nWe have to add in the sum(15-90) because in removing sum(3-99) and sum(5-100), we're taking out multiples of 15 twice.  Adding sum(15-90) ensures that we only take out 15 once, 30 once, etc.
809670
5 + 5 + 5 ≠ 550\n\nI'm not sure if that was the solution you were looking for though.\n\nEdit: nice job sunset!
88776
Are you looking for a set of 8 consecutive whole numbers that when you add 7 of them together in different combinations you receive those 3-digit responses?\n\nWell you've only listed 6 possibilities and only one of those is odd.\n\nStarting at 29 the lowest combination is 224 and the highest is 231.  Starting at 28 the lowest is 217.
1118957
The first number has to be (x+6) and the second number is (x).  There are four possible combinations: 6,0 (60); 7,1 (71); 8,2 (82); 9,3 (93).  60 plus 8 (68) does not equal 06, or 6.  71 plus 8 does not equal 17.  82 plus 8 does not equal 28.  93 plus 8 does not equal 39.\nUnfortunately, this problem is impossible.\nSorry I couldn't be of more help.
748483
The answer is 119\n\nTake 2 at a time end up with 1 (59 times=118 one left)\nTake 3 at a time end up with 2 (39 times=117 two left)\nTake 4 at a time end up with 3 (29times =116 three left)\nTake 5 at a time end up with 4 (23times=115 four left) \nTake 6 at a time end up with 5 (19times=114 five left)\nTake 7 at a time end up with 0 (17times=119 o left)\n\nI am dead sure this time = ]\n\nThe trick is look at 5 at a time. there is 4 left this means the last digit must be 9. So look at all possible answer of 7s that ends with 9 and work it out from that!!!!!!!!!!
674697
For five consecutive, there are 34 possibilities because you could start with 1, 2, 3,..., or 34\nYou can't start with 35 because then you would have\n35, 36, 37, 38 which is only four numbers.\nI'm not sure about the other two.\nNote: The answer to the other two are not 35 and 36.
920768
One solution is\n0, 2, 4, 6, 8, 10, 12, 1, 3, 5, 7, 9, 11\nBut there is a huge variety of solutions\n\nSorry Chris I did not copy so here is another\n\n0, 11, 5, 2, 9, 3, 1, 4, 10, 7, 12, 6, 8
875039
The first five are simple\nP(5,1) + P(5,2) + P(5,3) + P(5,4) + P(5,5) \n5 + 20 + 60 + 120 + 120 = 325\n\nSix and seven are a little more complicated\n6!/2! + 7!/2!2!\n360 + 1260 = 1620\n\nEight is special because you limit the maximum.\n\nList of bad values\n9x,xxx,xxx\n7x,xxx,xxx\n69,xxx,xxx\n67,xxx,xxx\n66,xxx,xxx\n65,9xx,xxx\n65,7xx,xxx\n65,6xx,xxx\n\n8!/3!2! - (2 * 7!/3!2!) - (3 * 6!/3!) - (3 * 5!/3!)\n3360 - 840 - 360 - 60 = 2100\n\nNow add the results.\n325 + 1620 + 2100 = 4045
306637
There's a trick to figuring this out quickly.\n\nFor each number between 1 and 499 there is a corresponding number between 501 and 999 which, when added together, equals 1000.\n\nFor example:\n\n1 + 999 = 1000\n2 + 998 = 1000\n...\n499 + 501 = 1000\n\nNow, there are 499 pairs like that.  Add them together and you get 499,000\n\nAdd to that 1000 and 500, and you get 500,500, which is your total.