# 0-9 digits how many four digit combinations can you have?

# How many different 4 digit combinations can be made using the digits 0 to 9?

Well you have 10 possible numbers for the fist column(0-9) and 10 for the second, third, and fourth then you multiply those numbers. 10*10*10*10=10000 or 10^4=10,000. So there are 10,000 different combinations.

# How many combinations can you make using four digits?

1234567890 Those are all the digits that exist. So, since there are 10 kinds of digits and four available spots, you must 10 4 . 10 times 10 times 10 times 10 = 10,000. There are 10,000 different combinations. The last person was too vague. I deleted his answer. Or, you can see tha…t the biggest number available is 9999, so 0-9999. 9999-0=9999. Add one because it's one of those, "how many are there" questions. you get 10,000 (MORE)

# How many numbers with exactly four digits can you make with the digits 012345678 and 9 the first digit may not be 0?

well you would probably have to mutiply 9 times ten times ten times ten, because you have four spaces for any number up to nine and zero can't be used in the first digit. My answer would have to be 9000, although I'm only in eighth grade. Anyone who has time to write out every possibility, though, h…as no life. (MORE)

# Numbers 0-9 how many different 13 digit combination can be made?

10,000,000,000,000.
This would be the number of permutations on a 13-position combination lock, for example..
In the first position there would be one number out of ten possibilities (0-9)..
Next to any one number there would be any one of ten again. So in the first two positions there are 10 x 1…0 possible combinations i.e. 10 ^2 (= 100). In the third position there would be 10 more possibilities against the previous 100 possibilities, i.e. 100 x 10, that is: 10 ^3 (= 1000). So, for a thirteen digit lock, there would be 10^13 possibilities, which is 1 followed by thirteen zeros, as shown above..
Of course, if 'numbers' with leading digits e.g. 0 000 087 654 321, and all zeros, are not counted as valid 'numbers' then there are 13 less possibilities. i.e. There are only 9,999,999,999,987 possible ' numbers'. (MORE)

# If you have 9 numbers how many different 6 digit combinations?

We treat this as a permutations and combinations problem... as there are two possible 'rules' involved. If we are allowed to repeat a digit (such as 456537 - which as two 5s in it) then there are 9 options for every digit. So 9 x 9 x 9 x 9 x 9 x 9 = 531441 different six-digit numbers from a c…hoice of 9. If we cannot repeat a digit, as we did above, thee need to pick the first number - and it can be any of the 9.. then the second number can be any of 8, and the third can be any of 7 and so on.... this is called permutations and we end up with a multiplication like before but with each figure lower than the one before it.... 9 * 8 * 7 * 6 * 5 * 4 = 60480 different permutations of six digits from a choice of 9. (MORE)

# What are the four digit combinations for 0-9?

If you don't mind starting off with one or more zeros, then they are every number from 0000 to 9999. That's 10,000 of them. If a number can't start with zeros, then it's every number from 1000 to 9999. That's the same 10,000 with the first 1,000 discarded ... total of 9,000 left.

# If you have twenty four digits how many combinations of twelve can you make?

If you have 24 distinct characters, then there are 2,704,156 combinations of 12 characters. Normally there are only ten digits and so 24 of them would contain duplicates. In that case the answer will depend on the duplication.

# What are all four-digit combinations of 0 to 9?

To be honest, I'm looking for the same thing - there's just too many! Sorry.

# How many 3 digit combinations are there in the numbers 1 to 9?

It depends. If you can only use each number once, there are 720 combinations. If you can use numbers multiple times, then there are 1000 combinations, by using all numbers from 000 to 999.

# How many combinations of the numbers 1 and 4 can there be using four digits?

Think of either using or not using the numbers 1 and 4 for each digit..
For example, let ____ ____ _____ ____ be the 4 digit number, now 1 can either go in to the first spot or 4 can. So there are 2 choices for that spot, and similarly two for the next spot and the next and the last. So there are 2…^4 =16 combinations of 1 and 4 if we make 4 digit numbers. (this uses the multiplication rule).
To help a little more to see it, look at just the two digit number, It should have 2^2=4 choices by the logic above..
Here they are:.
11,14,41,44 and that's it!..
See how we had two choices for each digit? (MORE)

# How many 9 digit combinations can be made from the numbers 123456789?

If the numbers are not allowed to repeat themselves then 362,880. * * * * * That is the number of permutations, not combinations. In a combination the order of the digits does not matter. So there is only one 9-digit combination.

# How many combinations are there for a 6 digit number using the numbers 0 to 9?

There are 1 million possible combinations. Just think of it as a sequence of numbers, starting at 000000, 000001... all the way to 999999.

# How many combinations of 9 digit numbers can be produced from the numbers 0 through 9?

If all numbers can be used as many times as wanted then there are 10 9 = one billion combinations. If each number can be used only once, there are 10!/(10-9)! = 10!/1! = 10! = 3628800 combinations. * * * * * Clearly answered by someone who does not know the difference between PERMUTATIONS and …COMBINATIONS. The combination 123456789 is the same as the combination 213456789 etc. All in all, therefore, there are only ten combinations which use each digit at most once. (MORE)

# How many combinations of three digits can be made from the numbers 0-9?

The question does NOT say that the three places must be differentdigits, so repetition is allowed. The first digit can't be zero, sothere are 9 choices for it. The second digit can be any one of the 10, including zero. The third digit can also be any one of the 10, including zero. Sothe quantity of …3-digit numbers that can be written is (9 x 10 x10) = 900 . (This makes sense. The smallest number youcan write with 3 digits is 100, and the largest is 999, which is atotal of 900 different numbers.) If repetition is not allowed, then: The first digit can't be zero,so there are 9 choices for it. For each choice, the second digit can be any one of the remaining 9(including zero). For each choice, the third digit can be any one of the remaining 8(including zero). So the quantity of 3-digit numbers that can bewritten without repetition or a leading zero is (9 x 9 x 8 ) = 648 . But the question doesn't ask "how many numbers can be made". It asks "How many combinations can be made ?" When you say"combinations" in math, that ordinarily means that the orderdoesn't matter, i.e. 247 would be considered the same thing as 724.Granted, this is getting pretty picky, but that's exactly what mathis all about, so we need to consider the possibility that thequestion was actually referring to combinations , and theorder of the digits doesn't matter. If repetition is not allowed:The first digit can be any one of 10, including zero. For each choice, the second digit can be any one of the remaining9. For each choice, the third digit can be any one of the remaining 8. The total number of 3-digit groups is (10 x 9 x 8) = 720. Each group of 3 digits can be arranged in (3 x 2) = 6 ways. So ifthe order doesn't matter, there are (720/6) = 120 separate, distinct COMBINATIONS of 3 digits. (MORE)

# How many four digit combnation can be made from 0-9?

Assuming you don't want a 0 as the leading digit, there are 9000 combinations, or 9*10*10*10. If you don't mind 0 as a leading digit, then 10000 combinations, or 10*10*10*10.

# How many 4 digit combinations can be made from the numbers 0 to 9?

If every number can be used as many times as you like, there are 10 4 = 10000 different combinations. If each number can only be used once, there are 9!/(9 - 4)! = 5040 combinations.

# List of four digit combinations of 0123456789?

The list is so easy to construct that you can even do it yourself. Here's how: -- Begin with the number 1,000. -- Count from 1,000 to 9,999. -- Write down every number you say. There's your list.

# How many 3 digit combinations can be made from 0 to 9 digits?

Assuming the digits can be used again, there are 10*10*10 or 1000 three digit combinations. IF YOUR LOOKING FOR 3 DIGIT LOTTERY ANSWERS HERE THEY ARE! Of the 1000 3-digit numbers 720 are non-repeating, 270 one-repeating, and 10 triples. Since a non-repeating number has 5 more siblings, we can divi…de 720 by 6 to obtain 120 distinct non-repeating-digit numbers which are listed below. 3-digit, non-repeating combinations (6-way) 012, 013, 014, 015, 016, 017, 018, 019, 023, 024, 025, 026, 027, 028, 029, 034, 035, 036, 037, 038, 039, 045, 046, 047, 048, 049, 056, 057, 058, 059, 067, 068, 069, 078, 079, 089, 123, 124, 125, 126, 127, 128, 129, 134, 135, 136, 137, 138, 139, 145, 146, 147, 148, 149, 156, 157, 158, 159, 167, 168, 169, 178, 179, 189, 234, 235, 236, 237, 238, 239, 245, 246, 247, 248, 249, 256, 257, 258, 259, 267, 268, 269, 278, 279, 289, 345, 346, 347, 348, 349, 356, 357, 358, 359, 367, 368, 369, 378, 379, 389, 456, 457, 458, 459, 467, 468, 469, 478, 479, 489, 567, 568, 569, 578, 579, 589, 678, 679, 689, 789.
Similarly, since three one-repeating numbers can be represented by one number, there are 270/3=90 distinct one-repeating numbers (doubles) as listed below. 3-digit, one-repeating combinations (Doubles) (3-way) 001, 002, 003, 004, 005, 006, 007, 008, 009, 011, 022, 033, 044, 055, 066, 077, 088, 099, 112, 113, 114, 115, 116, 117, 118, 119, 122, 133, 144, 155, 166, 177, 188, 199, 223, 224, 225, 226, 227, 228, 229, 233, 244, 255, 266, 277, 288, 299, 334, 335, 336, 337, 338, 339, 344, 355, 366, 377, 388, 399, 445, 446, 447, 448, 449, 455, 466, 477, 488, 499, 556, 557, 558, 559, 566, 577, 588, 599, 667, 668, 669, 677, 688, 699, 778, 779, 788, 799, 889, 899.
The following table and figures summarize the foregoing facts and the probability of winning with any order arrangement of 3-digit games. Group name and aliases How many 3-dig Numbers fall into this group How many distinct members of the group odds of winning with a number in the group Probability of a number drawn to be in the group .
Non-repeat (6-way) .
720.
120.
1:167.
72%.
One-repeat (3-way) (Doubles) .
270.
90.
1:333.
27%.
Triples .
10.
10.
1:1000.
1%.
One thousand if you allow repeating digits. (MORE)

# How many different combinations of three digits can be made from 0-9?

Only 9 numbers can be used in the hundreds place, and all the 10 numbers can be used for the tens and units place. So we have 9 x 10 x 10 = 900 three-digit numbers.

# How many different 4 digits combinations is there between 0-9?

Since 0 cannot be in the first place of the 4-digits number, we have 9 x 10 x 10 x 10 = 9,000 combinations.

# How many combinations of 9 digits are possible if no two adjacent digits are the same?

The first can be any one of 10 digits. For each of those . . . The second can be any one of the nine digits not the same as the first. For each of those . . . The third can be any one of the nine digits not the same as the second. For each of those . . . The fourth can be any one of the nine d…igits not the same as the third. For each of those . . . The fifth can be any one of the nine digits not the same as the fourth. For each of those . . . The sixth can be any one of the nine digits not the same as the fifth. For each of those . . . The seventh can be any one of the nine digits not the same as the sixth. For each of those . . . The eighth can be any one of the nine digits not the same as the seventh. For each of those . . . The ninth can be any one of the nine digits not the same as the eighth. The total number of possibilities is: (10 x 9 x 9 x 9 x 9 x 9 x 9 x 9 x 9) = 430,467,210 . If matching adjacent digits were allowed there would be 1,000,000,000 possibilities. So the restriction has eliminated 56.95% of them. (MORE)

# How many 4 digit combinations can be made with 5 digits?

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.

# How many 3 digit combinations can be made from 0 to 9 digits not repeating?

The correct answer is 720. This is obtained by multiplying the binomial coefficients C(10,1), C(9,1) and C(8,1). However, if you would not like the first digit to be a zero, then the answer changes slightly. There are now 648 solutions. This is found by counting all 3 digit combinations in which zer…o is last, second, and not involved. (MORE)

# How many different 3 digit combinations can there be if the digits are distinct?

With 3 digit numbers there are 10 digits (0, 1, ..., 9) from which to select 3 different digits. This can be done in: 10 C 3 = 10! / (10-3)!3! = 120 ways. .
For a set with n items, r of them can be selected (permuted) in: n x (n-1) x ... x (n - (r +1)) = n! / (n-r)! = n P r .
ways. Howe…ver, if the order of selection does not matter (just the combination of the r items) then the selected items can be chosen in r x (r-1) x ... x 1 = r! .
ways. Meaning there are n P r / r! = n C r = n! / (n-r)!r! .
combinations possible. (MORE)

# How many number combinations for 9 digits?

999,999,999 if you don't include 000000000, otherwise 1,000,000,000. If you don't allow repeated digits then 9 factorial, ie 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 = 362,880.

# How many 9 digit combinations are there for numbers 0 through 9?

That's exactly all the numbers you need to count from 000,000,000 to 999,999,999. There are one billion of them.

# How many possible combinations are there using 6 digits from 0 to 9 beginning with any digit and repeating any digit ie commencing with 000000 and ending with 999999?

That's simple enough. Just count the numbers from 0 to 999,999 and you will find there are one million of them.

# How many 5 digit combination are there 0-9?

infinite. not kidding. * * * * * Not kidding but NOT CORRECT either! There are 10!/(5!*5!) combinations where n! = 1*2* ... *n 10*9*8*7*6/(5*4*3*2*1) = 252 combinations.

# How many 3 digit combinations can be made with 4 digits?

nCr = n!/r!(n-r)! = 4 I'm using the combination formula where the place of the digits is not important No if the place for instance the combination 312 is the same as 213. But I think you are asking for a permutation instead where the placements of the digits are also important. So it would… actually be 4!/(4-3)! or 24. In a permutation the place is significant i.e. 312 and 213 are distinct even though they have the same 3 digits. (MORE)

# What are all of the combinations of the numbers 0-9 for a four digit number?

There are (10*9*8*7)/(4*3*2*1) = 210 combinations. And I certainly am not inclined to list them all. However, if you are methodical about it, the job will be time consuming but not hard. Perhaps you might start with 999_ and list all the choices for _ then 998_ and list all those choices … etc. (MORE)

# How many 3 digit combinations can be made from 1 to 9?

All the numbers from 0 to 999 so 1000 numbers However if you mean only using each number once the answer is about 720 * * * * * No. The above answer refers to the number of permutations. Permutations are NOT the same as combinations as anyone who has studies any probability theory can tell y…ou. The number of combinations is 9 C 3 = 9*8*7/3*2*1 = 84 (MORE)

# How many number combinations for 9 digits repeat allowed?

More then 30 how do i know that because i did it it took me more then 3 hours * * * * * Given that in a combination, 123124125 is the same as 111222345 or 321521421 etc (the order does not matter), I make the answer 8,697,700 - anyone willing to check? This is much smaller than the 900,000,0…00 nine-digit numbers that do not have leading zeros, or the 999,999,999 nine-digit number including leading zeros. (MORE)

# How many 2 digit combinations can be made with 99 digits?

45. There are only 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. So I am not sure where the "99 digits", in the question come from.

# How many 2 digit combinations can be made from 0 to 9 digits?

45 In combinations, the order of the digits does not matter so that 12 and 21 are considered the same.

# How many four digit numbers can be created using the digits 0 through 9?

If leading zeros are allowed, then you have 0000 through 9999 which is 10,000 numbers . .
If leading zeros not allowed: the smallest four-digit number is 1000, eliminating 0000 through 0999 (one thousand numbers are not allowed), so it is 9,000 numbers .

# How many combinations of four digits can be made from 1-44?

There are 135751 combinations of 4 numbers (NOT DIGITS) that can be made from the numbers 1 to 44.

# How many combinations are there in a 9 digit number?

Does order matter? A common mistake is saying "combinations" when you actually mean "permutations". In the topic of Probability, permutations are ordered , while combinations are unordered . This means that for permutations, "123456789" and "987654321" are each considered unique results. W…hen dealing with combinations, they are considered the same result . If you actually meant how many permutations are there, then the answer is as follows: b 9 , where b is the base in which the number is expressed. In decimal, that would be 10 9 , or 1000000000 d . In binary, that would be 2 9 , or 512 d . However, if you truly meant to ask how many combinations exist for a 9 digit number, (assuming base 10 using digits 0 thru 9) the answer is: 10. The formula for combinations is spoken as "n choose r", or n C r , where n is the number of available things to choose from, and r is how many things you are choosing from what is available. The combination formula is: n! / (r! * (n - r)!) So for your question, the equation would be as follows: 10! / (9! * (10-9)!) = 3628800 / (362880 * (1!)) = 3628800/362880 = 10 There are 10 combinations available for a 9 digit number when utilizing the digits 0 thru 9. Listing them out is as simple as excluding one of the digits from each combination: 123456789 (excludes 0) 012345678 (excludes 9) 012345679 (excludes 8) 012345689 (excludes 7) 012345789 (excludes 6) 012346789 (excludes 5) 012356789 (excludes 4) 012456789 (excludes 3) 013456789 (excludes 2) 023456789 (excludes 1) (MORE)

# How many 5 digit combinations can be made from 9 digits?

405 combos. * * * * * That may be so in a galaxy far far away, but in the real world, the answer is 9 C 5 = 9*8*7*6/(4*3*2*1) = 126 combinations.

# How many 4 digit combinations can be made with 3 digits?

27since there are 3 choices for the first digit and then 3 for the second and 3 for the third which is 3x3x3=27

# How many 3 digit combinations there 0-9?

The order of the digits in a combination does not matter. So 123 is the same as 132 or 312 etc. There are 10 combinations using just one of the digits (3 times). There are 90 combinations using 2 digits (1 once and 1 twice). There are 120 combinations using three different digit. 220 in all.

# How many 5 digit combinations can be made from 0 to 9 digits?

You don't say whether you can repeat a digit. Assuming you can - it's 10 x 10 x 10 x10 x 10 which is 10,000.

# How can there be in the many 7 combinations can there be in the digits 1 through 9?

If you mean how many different sets of 7 can be made from 9 numbers, the answer is (9 x 8)/2 ie 36.

# How many four digit combinations are there in the numbers 1 to 12?

There are 12 C 4 4 NUMBER combinations. And that equals 12*11*10*9/(4/3/2/1) = 495 combinations. However, some of these, although 4 number combinations consist of 7 digits eg 1, 10, 11, and 12. Are you really sure you want 4-DIGIT combinations?

# How many different combinations of 3 can be made from the digits of 0-9 if no digits are repeated?

If the sequence of the digits is important ... 1-2-3 and 3-2-1 are different ... then there are 10*9*8 = 720 of them. If the sequence doesn't matter ... 1-2-3 and 2-3-1 and 3-1-2 are all the same ... then there are only 720/6 = 120 because 3 things can be ordered 1*2*3 = 6 ways.

# How combination in a 9 digit number?

There are different numbers of combinations for groups of different sizes out of 9: 1 combination of 9 digits 9 combinations of 1 digit and of 8 digits 36 combinations of 2 digits and of 7 digits 84 combinations of 3 digits and of 6 digits 126 combinations of 4 digits and of 5 digits 2…55 combinations in all. (MORE)

# How many four digit numbers can be made using the digits 4 3 0 and 9 without repeating the digits?

The general rule is to imagine that you have four places in the number eg 9999 then calculate by saying 4 x 3 x 2 x 1 = 24. That gives you: 24 altogether if you allow numbers starting with ZERO. 18 otherwise. SIX starting with 0: 0349 0394 0439 0493 0934 0943 SIX starting with 3: … 3049 3094 3409 3490 3904 3940 SIX starting with 4: 4039 4093 4309 4390 4903 4930 SIX starting with 9: 9034 9043 9304 9340 9403 9430 (MORE)

# How many 3 digit combinations can be made from 0 to 6 digits?

I'll assume that you mean " ... from the digits 0, 1, 2, 3, 4, 5,and 6 ?" . -- If the same digit can appear more than once in the samecombination, then there are (7 x 7 x 7) = 343 possibilities. -- If the three digits in one combination must all be differentdigits, then there are (7 x 6 x 5) = 210… possibilities. -- If the combinations are actual cou nting numbers, and so thefirst digit can't be 'zero', then there are (6 x 6 x 5) = 180 possibilities. (MORE)

# How many three digit number combinations are there in 0 to 9 if you can not repeat a number?

This question is ambiguous. It could mean the numbers 123,456,789 in which case the answer is three as only three combinations of these digits can be used whether or not the 0 is used. Or more likely it could mean the range 100, 101,102,103 etc up to 999 in which case you can work out what the answe…r is, by subtracting the first number in this range from the last number. (MORE)

# How many 4 digit combinations are there for numbers 0 to 9?

-- If you're allowed to repeat digits in one combination, then there are 10 4 = 10,000possibilities . -- If all four of the digits must be different, then there are (10 x 9 x 8 x 7) = 5,040 possibilities .

# How many 9 digit combinations using 0 through 9?

There are ten combinations: one each where one of the ten digits, 0-9, is excluded.