NOT AN ANSWER:
More acceptable combinations:
1231 3213 2321
More unacceptable combinations:
1112 2222 3113
Thanks.
The fraction of 0.11111 (Just assuming you meant repetent by the repeated '1's) is 1/9.
Note that the numbers from 6000 to 6999 comprise all combinations of three digits, with a 6 stuck on the front. We will consider the combinations of these final three digits.First, let's consider how many three-digit combinations contain one, two or three 7's.There is one occurrence of three 7's: in the combination '777'.Now consider two 7's. This can occur in the pattern '77x', '7x7' or 'x77'. For each of these patterns there are 9 possible values for 'x'. Why not 10? Because if x were '7', then it would be '777' which has three, not two 7's. So there are 9 times 3, or 27 ways for there to be exactly two 7's.Now consider how many occurrences of one 7. The three patterns are '7xy', 'x7y' and 'xy7'. There are 81 combinations of x and y for each of these. (Why not 100? because we are excluding 19 cases where x or y, or both, are 7's.) So there are 81*3, or 243 ways that there can be one 7.So there are 1 + 27, or 28 ways for there to be at least two 7's. And there are 28 ways for there to be a repeated 8, or a repeated 9, and so on. So there are 9 times 28, or 252 ways for a digit other than 6 to be repeated. Why exclude 6? Because we're going to treat it specially.All the numbers from 6000 to 6999 begin with a 6. So it is only necessary for a single 6 to occur in the last three digits for it to repeat. And there are 243 + 27 + 1, or 271 ways that one or more 6's can occur, as we saw above.But wait, we cannot just add 271 to 252, because we would be counting combinations like '6677' or '6767' twice. In fact there are 9 times 3 cases with a repeated 6 and some other repeated digit. (three patterns, and nine possible other digits) So the final count will be 271 + 252 - 27, or 496.So there are 496 numbers between 6000 and 6999 that have at least one repeated digit.
The repeated factor of 102 is 10.
The repeated prime factor is 2.
It is 33.33... repeated %.
Assuming the digits are not repeated, there are four combinations:123, 124, 134 and 234.
Assuming that repeated numbers are allowed, the number of possible combinations is given by 40 * 40 * 40 = 64000.If repeated numbers are not allowed, the number of possible combinations is given by 40 * 39 * 38 = 59280.
Assuming no repeated digits, lowest first, 20; in any order 120; Allowing repeated digits: 216
If the 6 digits can be repeated, there are 1296 different combinations. If you cannot repeat digits in the combination there are 360 different combinations. * * * * * No. That is the number of PERMUTATIONS, not COMBINATIONS. If you have 6 different digits, you can make only 15 4-digit combinations from them.
If the numbers can be repeated and the numbers are 0-9 then there are 1000 different combinations.
Assuming the digits cannot be repeated, there are 7 combinations with 1 digit, 21 combinations with 2 digits, 35 combinations with 3 digits, 35 combinations with 4 digits, 21 combinations with 5 digits, 7 combinations with 6 digits and 1 combinations with 7 digits. That makes a total of 2^7 - 1 = 127: too many for me to list. If digits can be repeated, there are infinitely many combinations.
If you allow digits to be repeated (for example, 222 or 992), then there are 9 x 9 x 9 = 729 combinations. If you do not allow digits to be repeated, then there are 9 x 8 x 7 = 504 combinations.
Letters - 26Numbers - 10Total chacters - 36If characters can be repeated (QQ and 44 are acceptable): 36 x 36 = 1,296If characters can't be repeated (GG and 99 not acceptable): 36 x 35 = 1,260If characters can't be repeated and the sequencedoesn't matter (X7 and 7X are considered to be the same): 630
10 to the third assuming zero is included and can lead, otherwise 9 to the third, ie 729. This all assumes that you are talking about 3-digit numbers. * * * * * No. That may be the number of permutations but those are different from combinations. In a combiation, the order of the digits does not matter so that 123 is the same as 132 or 213 etc. With repetition, there are 210 COMBINATIONS, including one that is {0,0,0}.
If you have 4 positions, each of which can hold any of the ten digits, you have 10 to the power 4 combinations. If you can have only 4 different digits, you have 4 to the power 4 different combinations.
If no digit can be repeated then there are 5 combinations, abcd, abce, abde, acde and bcde. If you regard abdc as different from abcd then each of the 5 basic sets could be arranged 24 ways and the total would be 120 combinations.
be repeated until performace is at an acceptable level