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If the numbers can be repeated and the numbers are 0-9 then there are 1000 different combinations.

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Q: How many number combinations are possible with three numbers?

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Six * * * * * No, that is the number of PERMUTATIONS (not combinations). With 3 numbers, the number of combinations, including the null combination, is 23 = 8. With the three numbers 1,2 and 3, these would be {None of them}, {1), {2), {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.

There are an infinite number of possibilities based on the infinite set of numbers. However, for a finite set, there are limited possible combinations, depending on whether you can use the same numbers over again, or if they have to be distinct, or if their order makes any difference. Here's an example: For a group of THREE numbers, there is only one possible group of 3 numbers, and there are three possible groups of 2 numbers (i.e. 12, 13, 23) . Using each of three different numbers, there are 6 ordered combinations of two numbers (12, 13, 23, 21, 31, 32) and 6 possible combinations of three numbers (123, 132, 213, 231, 312, 321). If the numbers are allowed to repeat, there are 9 possible combinations of two (add 11, 22, 33) and 8 more possible combinations of three (111, 112, 113, 122, 133, 222, 223, 333) - if order matters, each triple (111) has only one possible order, each double has three (112, 121, 211). The number rapidly increases for larger numbers of possible and larger groups from those sets. The possibilities are called combinations and permutations, and are connected to the numerical property called "factorials" (a number multiplied by all smaller integers - 2 factorial is represented by "2!" and equals 2 x 1 = 2, while 3! = 3 x 2 x 1 = 6). The number of discrete sets of K numbers from N possible numbers is N! / K! x (N-K)!

If you have 12 possible numbers with multiple combinations then you should start out with making all the possible combinations; you will find theyre 20. Theyre four numbers out of the twleve that can be divisible by three; 3, 6, 9, and 12. There are 7 combinations where the combinations can equal those four numbers. So the odds of getting a sum divisible by three is 7/20.

There is 1 combination of all ten numbers, 10 combinations of one number and of nine numbers, 45 combinations of two or eight numbers, 120 combinations of three or seven numbers, 210 combinations of four or six numbers and 252 combinations of five numbers. That is 1023 = 210 - 1 in total.

Three possible combinations: 17+1, 13+5 and 7+11.

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Six * * * * * No, that is the number of PERMUTATIONS (not combinations). With 3 numbers, the number of combinations, including the null combination, is 23 = 8. With the three numbers 1,2 and 3, these would be {None of them}, {1), {2), {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.

There are an infinite number of possibilities based on the infinite set of numbers. However, for a finite set, there are limited possible combinations, depending on whether you can use the same numbers over again, or if they have to be distinct, or if their order makes any difference. Here's an example: For a group of THREE numbers, there is only one possible group of 3 numbers, and there are three possible groups of 2 numbers (i.e. 12, 13, 23) . Using each of three different numbers, there are 6 ordered combinations of two numbers (12, 13, 23, 21, 31, 32) and 6 possible combinations of three numbers (123, 132, 213, 231, 312, 321). If the numbers are allowed to repeat, there are 9 possible combinations of two (add 11, 22, 33) and 8 more possible combinations of three (111, 112, 113, 122, 133, 222, 223, 333) - if order matters, each triple (111) has only one possible order, each double has three (112, 121, 211). The number rapidly increases for larger numbers of possible and larger groups from those sets. The possibilities are called combinations and permutations, and are connected to the numerical property called "factorials" (a number multiplied by all smaller integers - 2 factorial is represented by "2!" and equals 2 x 1 = 2, while 3! = 3 x 2 x 1 = 6). The number of discrete sets of K numbers from N possible numbers is N! / K! x (N-K)!

If you have 12 possible numbers with multiple combinations then you should start out with making all the possible combinations; you will find theyre 20. Theyre four numbers out of the twleve that can be divisible by three; 3, 6, 9, and 12. There are 7 combinations where the combinations can equal those four numbers. So the odds of getting a sum divisible by three is 7/20.

I'm assuming they're three unique numbers. Thus, the first can be any of three, the second either of the remaining two, and the last is the last one left. Thus: combinations = 3 * 2 * 1 = 6 Or, more generally, the combinations of n numbers in such a problem is n factorial, denoted as "n!", which is every number from 1 to that number multiplied together.

There is 1 combination of all ten numbers, 10 combinations of one number and of nine numbers, 45 combinations of two or eight numbers, 120 combinations of three or seven numbers, 210 combinations of four or six numbers and 252 combinations of five numbers. That is 1023 = 210 - 1 in total.

Three possible combinations: 17+1, 13+5 and 7+11.

Working out the combination of numbersThe number of three-digit integersAssumptions: You are using the numbers 0-9 (10 combinations each)You can use any number more than onceYou include leading zeros, ie. 001, 002'000' is a possible combinationThere are exactly 1000 possible ways if all assumptions are true, and 900 if numbers under 100 do not count.How to work it outMultiply the possible combinations so that it will be 10*10*10 or 103 (three times because there are three slots)

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If the numbers contain zeros, the total number of combinations is 10,000. You can work this out easily logically: For ten single-digit numbers (0,1,2,3,4,5,6,7,8,9) then there are 10 possible 'combinations' For numbers with 2 digits then for each possible digit in the 10s column (e.g. in the 20s range) there are another 10 possible combinations (20,21,22,23,24,25,26,27, 28,29). As there are 10 possible ranges (single digits, teens, twenties, thirties etc) there will be 10 X 10 or 100 possible combinations. using the same logic, for three digits, there will be 10 X 10 X 10 or 1000 digits. And for 4 digits there will be 10 x 10 x 10 x 10 = 10,000 possible combinations. So for a number, say, with x digits, the total number of combinations of those digits will be 10 x 10 x 10..... etc with x numbers of 10s in the calculation. You can find out the number of combinations of any set of letters or numbers in the same way. as an example, to find out, say, the possible combinations of letters in the alphabet of 26 letters, then using the same method this can be given as 26 x 26 x 26 x 26............. with 26 '26's' in a row multiplied together. This gives the staggering amount of approximately 615612 followed by 31 zeros.

There are 18*17*16/6 = 816 of them!

There are 60 possible numbers for the first number, A, in the combination (1,2,3,...,59,60).For each of outcome of A, there are 60 outcomes for the second number, B, giving a two digit combination 60x60=360 possibilities.For each of these 360 outcomes, there are 60 outcomes for the third number, C, making the number of possible combinations 360x60=21600.Or 60 possibilities x 60 possibilities x 60 possibilities = 21,600 possibilities.

Start with the factors. Multiply combinations of three prime factors, then combinations of five, then seven, etc. All generated numbers will be guaranteed to have an odd number of prime factors.