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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 divide 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 aliasesHow many 3-dig Numbers fall into this groupHow many distinct members of the groupodds of winning with a number in the groupProbability of a number drawn to be in the groupNon-repeat (6-way)7201201:16772%One-repeat (3-way) (Doubles)270901:33327%Triples10101:10001%

One thousand if you allow repeating digits.

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

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9,000 - all the numbers between 1,000 and 9,999 inclusive. * * * * * NO. Those are PERMUTATIONS, not COMBINATIONS. Also, the question specified 4 digit combinations using 4 digits. The above answer uses 10 digits. If you start with 4 digits, you can make only 1 combination.

Only one: 2468. The order of the digits in a combination does not make a difference.

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.

There are 9C3 = 84 combinations.

Using the eight digits, 1 - 8 ,-- There are 40,320 eight-digit permutations.-- There is 1 eight-digit combination.

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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.

Only one.

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

about 1,0000000000000

9,000 - all the numbers between 1,000 and 9,999 inclusive. * * * * * NO. Those are PERMUTATIONS, not COMBINATIONS. Also, the question specified 4 digit combinations using 4 digits. The above answer uses 10 digits. If you start with 4 digits, you can make only 1 combination.

120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.

Only one: 2468. The order of the digits in a combination does not make a difference.

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.

10C6 = 10*9*8*7/(4*3*2*1) = 210 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.

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