how many 4 digit numbers are there if the amount of numbers do not contain 0 and a repeating number?
956128
Simply multiply\n9 * 8 * 7 * 6\nBecause the first number can be 1-9 but not zero and the second number can be 1-9 but one number has already been used and so on and so forth.\nHope it helps.
845925
If you problem, or part of it, is:\n\nHow many 8 digit numbers can be formed from this list of digits?\n8! / (3! * 2!)\nThe number combinations of digits has tobe reduced by 2! because of the 2 "6"s.\nThe number combinations of digits has tobe reduced by 3! because of the 3 "2"s\n.\nBut since the number cannot exceed 65,500; all larger numbers must be eliminated. Subtract the number of 8 digit numbers that can be produced from the list that start with "9". Likewise "7". Likewise "69", "67", "66", "659", ""657", and "656".\n\n\n\n(Try to state your problem clearly and completely.)
580639
6,643,342,716\n\nno wait, it's 6,643,342,791\nwait, 6,643,342,813\noh, 6,643,342,876\n\nwould you guys cut that out a minute...
215950
almost ... its 1000 to 9999 \n\ni.e 9000 combinations :)\n\nAnother way to do this ... \n\nexcept the first digit all can have a 0 .. so the possible combinations are : \n\n9*10*10*10 = 9000\n\n( 9 possible digits can be put in the first and 10 in the 2 thru 4)
959905
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!!
909218
if it has three or more digits, and if it ends in zero.
1169470
792\n\n1 thru 9 = 9 digits\n10 thru 99 = 180 digits (90 numbers with 2 digits each)\n100 thru 300 = 603 digits (201 numbers with 3 digits each)\nTOTAL = 792
886465
You need the 10 digits 0 through 9 (obviously). You also need another 1 and 2, so you can represent 11 and 22.
9961
0,1,1,2,3\n\nto get the next number add the previous 2 numbers
727389
1) 21\n2) 52\n3) 40\n4) 80\n5) 104\nIn each pattern you are adding a number to the one before it.\nEx: 1 + 0 = 1, 1 + 1= 2, 2 + 1 = 3, 3 + 2 =5, 5 + 3 = 8, etc.
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 :))
178354
if there are finitely many digits after the decimal point, count them, if there's n of them, write what you had behind the decimal point over 10^n, that is a 1 with n zeroes.\nIf there are infinitely many, find the moment when the digits begin repeating.  put the repeating sequence over as many nines as there are repeating digits.For a complex case, see the example (when there's a repeating and not repeating part.\nExample\n0.34356356356...=0.34+0.00356356356.... [notation is sometimes 0.00(356)]\n0.34=34/100=17/50\n0.00(356)=1/1000*0.(356)=1/1000*(356/999)=356/999000
177153
There is no flaw as long as you recognize that .99999 repeats forever.  If you were to end it at some point and subtract, then as a previous poster said, you'd actually get 8.99....1.  But since it repeats, there is always another 9 and so it works.\n\nAnd consider this - if .9999999repeating did NOT equal 1, then there would be another number between them.  What would that number be?\n\n~~~~~~~~~~~~~~~~~~~~\nNo, the number in question is not infinitessimal, it doesn't exist.  You can never get to the end of the string of nines to try and squeeze another number in there :)
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!!!!!!!!!!
125542
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.
359204
100 which is the binary equivalent of 4 if you are considering those 2 numbers to be added binary numbers
798343
The first 3 digits cannot be 800, or 900, or all the numbers from 000 to 099.  That leaves a total of 1000 - 102 = 898 possibilities left.   The next 3 digits can be anything from 000 to 999, or 1000 possibilities.  The last 4 digits must be 0000, or just one possibility.  The answer is the product: 898 x 1000 x 1, or 898,000.