I´ve asked how can I calculate de index of alphabet combination = 26 letters in groups of 3. I mean C(26,3).\nThe answers I´d received were very very useful.\n\nNow, I want to perform the opposit.\n\nIs it possible? Look at my previous question in order to understand this one.\n\nI have the number 1728 and I know it is the index of the alphabet combination C(26,3).\n\nHow can I reach to H-O-Y when I have only the index?\nI mean, how is the reversing performed?\nThanks a lot.
123103
the method is called double dabble, divide by 26 round subtract divide by 26 etc.
513529
I see too many combinations to make anything out easily.\nI thought maybe the first might be PR, could also be AXIO...(didn't quite make AXIS POWERS, too bad!)  All those dots in a row have me stumped.  Letters using only dots: e (1 dot), i (2 dots), s (3 dots) and h (4 dots), as well as the number 5 (5 dots).\nAny clues?
535461
I understand what you are saying when you mention the permutations for 10 digits, but I think you're missing what it really means.  First, this isn't right to think of it this way to think about this.  Non-repeating doesn't mean that I can't find a string in the decimal expansion that's the same as an earlier string (for example, finding a piece 14159 later in pi isn't impossible).  What it means is that there is no point where the digits in pi *completely* repeat (i.e. it becomes something like 141591415914159.... on and on forever).  So you're right, it can include some kind of partial repeat, but non-repeating means it can't have a total repeat.\n\nYour comment about the permutations (not combinations) on the numbers {0,...,9} is right but it's used incorrectly also.  It's wrong because of what I mentioned above.  Basically it's assuming that once you pick a number you can't repeat it again in a 10 digit sequence, but that's clearly wrong (again, 14159 appears in pi, where 1 occurs twice).  So you have several right ideas here but you're combining them seomwhat incorrectly.\n\nHope this clears things up!!
69777
I can get you started. \n\nThe digit A can't be very large if ABCD * 4 is a 4 digit number. In fact 4  3 = 12 so if A = 3 then ABCD x 4 is at least five digits. \n\nA is either 1 or 2. If A is 1 then 1BCD x 4 = DCB1 But this is not possible since the left side is even and the right side is odd. Thus A = 2. \n\nso, we have the first number!\n\n2BCD x 4 = DCB2 \n\nLook at D on the right side. What can it be? 4 x 2 = 8 so D might be 8 or, if there is a carry, D might be 9. But if D on the left side is 9 we get 4 x 9 = 36 which doesn't work with the units digit of 2 on the right side. So, D = 8 and we get:\n\n2BC8 x 4 = 8CB2 \n\nThis should be enough to get you started
417645
If the number begins with 5, there are still three of the odd digits available to fill the last place.  If a 3 is used for the last place, the others are filled using 1, 2, 3, 4, and so there are 4! arrangements.\nIf 1 is used for the last place, that leaves 2, 3, 3, 4 for the others, and so the number of arrangements is  4!/2! because of the repeated 3.  So that gives \n4! + 4!/2! = 36\n\nIf it begins with a 3, the last place can be filled with 1, 3, or 5, and  the numbers left to fill the remaining four places are all different and so the number of possibilities is\n3*4! =72\n\nIf it begins with a 2 or 4 ( two choices), and the last with 1 or 5 (two choices), there are 4!/2! ways of filling the other places because 3 is repeated.\n2*2*4!/2! = 48\n\nIf it begins with 2 or 4 and ends with 3, then the other four digits are all different, and the number of possibilities is\n2*4! = 48\n\nAdding all these together gives 204.  Yes.
520135
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.
227314
We have 3 products of single digit numbers (0-9) that all equate to the same value (call it x):\n\n1) abc = x\n2) bge = x\n3) def = x\n\nWe can't have one of the digits be 0 because its product would be 0.  To make the other products zero, one of their digits would also have to be zero, but no letter appears in all 3 equations.\n\nSimilarly, 5 cannot be one of the digits because that would require a five in all the other equations.  No other digit is divisible by 5 so there is no way to get a multiplier of 5 except from another 5.\n\nLikewise 7 cannot be one of the digits for the same reasons.  There are no other digits that are divisible by 7.\n\nThat leaves us with the set {1,2,3,4,6,8,9} for the seven letters a,b,c,d,e,f,g.  So we know the values of abcdefg, just need to figure out what equates to what.\n\nLet's think about the combinations for abc and def that will multiply to the same product:\n\nThe only combinations that work are:\n2x3x6 = 36\n1x4x9 = 36\n\nOr:\n3x4x6 = 72\n1x8x9 = 72\n\nIf we try to work with the first product (36), there is no way to form another product bge that is also 36.  Try it for yourself.\n\nHow else can we make 72 though?  Only one way.\n2x4x9 = 72\n\nThe one new digit used is 2.  So we are done.  We don't know the exact assignments of abc and def, but we do know that g must equal 2.\n\nThe products are:\n3x4x6 = 72\n1x8x9 = 72\n2x4x9 = 72\n\na is one of {1,3,6,8}\nb is one of {4,9}\nc is one of {1,3,6,8}\nd is one of {1,3,6,8}\ne is one of {4,9}\nf is one of {1,3,6,8}\ng is 2\n\nThe answer is g is 2.
72899
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/
514960
I'll explain how you do it but you should be able to do the thinking or you're going to find maths very hard.2,4,6,8,0-use as many different 3 digit combinations as you can.for eg.\n246,468,642,420,462,222,444,666,606...........get the idea?
196341
Okay, first you have to calculate how many the combinations of the first three teams are. At this point, arrangement is irrelevant. So, it'll be 15C3. Sorry, I don't exactly remember the full formula. Then, you'll get 455. Now, the arrangement is becoming relevant, since, the three teams could be number one, two, or three. So, at this point, permutation takes place. You are to multiply the previous answer with 3P3(or 3!, same things, aren't they?!). So,the answer will be 2,730! I know, it could be similar to another or other answer(s), but in math, the way of solving a question is more important than the answer itself! (That's what i think!)
521422
So is this sequence based purely on a mathematical function?\n\nOkay, so there's a 52 in there, could this sequence somehow relate to the calendar?\n\nWhat sort of scientist? I'm a computer scientist and nothing's glaringly obvious about those numbers.\n\n** GOT IT *** 68! Er or Erbium, the sequence spells out Brainteaser based on the atomic weights of B, Ra, In, Te, As, and Er.
49567
Let the last digit be n. The first digit is at least 1, so we need to count the number of possible partition of n-1 into 4 digits. There are (n-1+3)C(3) ways.\n\nThere are 4 possibilities for the last digit, 2, 4, 6, and 8. So the answer is (4)C(3) + (6)C(3) + (8)C(3) + (10)C(3) = 4 + 20 + 56 + 120 = 200.
557691
if you know permutation/combination..the answer is easy to understand...\n\nlike 23C23+23C22+23C21+......+23C1
103962
In the experiment of choosing the 3 digits\n\nTotal no. of possible choices \n= (no.of ways in which the first digit can be chosen) * (no.of ways in which the second digit can be chosen) * (no.of ways in which the third digit can be chosen)\n==> n = 9 * 10 * 10\n==> n = 900\n\n[Since the first digit cannot be zero, there would be 9 possibilities for choosing it, 10 for second and third]\n\nLet "A" be the event that all the 3 digits are different and "B' the event that the 3rd digit is different from the first 2 digits.\n\nFor Event "A"\n\nNo.of favorable choices\n= (no.of ways in which the first digit can be chosen) * (no.of ways in which the second digit can be chosen) * (no.of ways in which the third digit can be chosen)\n= 9 * 9 * 8\n==> m(a) = 648\n\n[Since the three digits to be chosen should be different, there would be 9 possibilities for choosing the first digit (since 0 is excluded), 9 possibilities for the second (excluding the one chosen for the first and including 0) and 8 for the third (excluding the ones chosen for the first and the second and including 0)\n\nProbability that all the digits are different\n==> Probabilty of occurance of event "A" = m(a)/n\n==> P(A) = 648/900\n==> P(A) = 18/25\n\nFor Event "B"\n\nNo.of favorable choices\n= (no.of ways in which the first digit can be chosen) * (no.of ways in which the second digit can be chosen) * (no.of ways in which the third digit can be chosen)\n= 9 * 10 * 8\n==> m(a) = 720\n\n[Since the first two digits can be any of the ten and the last should be different from the first two, there would be 9 possibilities for choosing the first digit (since 0 is excluded), 10 possibilities for the second (any digit can be chosen including 0) and 8 for the third (excluding the ones chosen for the first and the second)\n\nProbability that the 3rd digit is different from the first two\n==> Probabilty of occurance of event "B" = m(b)/n\n==> P(A) = 720/900\n==> P(A) = 4/5\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/
636892
There is definitely something wrong with the wording of the puzzle, because you can just turn over the [A] card. Anyways, I doubt your problem is that easy. I'm guess that you're talking about the reverse of both Aces.\n\nSince each card card will have only a number and a letter, there are only 4 possibilities:\n\n1)\nFront: [A] [K] [2] [7]\nBack: [2] [7] [A] [K]\n\n2)\nFront: [A] [K] [2] [7]\nBack: [7] [2] [A] [K]\n\n3)\nFront: [A] [K] [2] [7]\nBack: [2] [7] [K] [A]\n\n4)\nFront: [A] [K] [2] [7]\nBack: [7] [2] [K] [A]\n\nSince there are actually 2 Aces, you could play a trick question and flip over the other ace as well. So even if there is a 2 behind the [A], there might be another Ace behind the [2]. The easy way is to flip over the [A] and the [2].\n\nHowever, you can do it 3 other ways. By flipping over exactly 1 face card (A or K) and 1 numbered card (2 or 7), you can determine exactly what's behind all four cards.\n\nFor example, if you flip over the [7] and the [A]:\nIf a 7 is behind the [A], then the 2 is behind the [K], and vice versa. If the A is behind the [7], then the K is behind the [2], and vice versa.\n\nOnce the combination is found, you will know if either 2 is paired with either Ace. (If they are randomly sorted, the probability will be 3 out of 4).
514130
1275 may be the correct answer but not for that reason. i don't agree with the logic in finding the pattern. i have been trying to find the answer for the last 20 mins and will keep trying to find the answer but i want to know two things:\n\ndoes ab represent just the unknown number or is it a two digit number\n\nand are the numbers correct? \n\none more thing, how was this question written i.e. horizontally, vertically?\n\ncheers\n\nthe differences between each number starting from the bottom is \n4, 8, 200, 330 and the differences between those numbers are 4, 192, 132. so i can't get a guesstimate for you other than it should be around 1000 and probably ends in five.
542292
OMG a math riddle I cant seem to answer!!!! :-)  Awesome. I cant sleep because now i want to knwo the anwer. And so i continue to try and solve it.\n\nBest i can come up with is 113  or 240 or 50-something..\nAlso maybe another number in the 100-n-teens\n\nI did find a 1-3-4 pattern aasuming the next number in 1-3-4-1 was a 3 but was unsure of how to continue it. (its1230 in the moring here)\n\nAnd again another possible patter 1-3-4-8 next number would be 16.\n\nI know some may not even know how I came up with these, but I love math and I love a challenge. I have been trying to find any possible correlation in the numbers that I can. \n\nHope someone figures it out. I know I cant wait to see the answer.