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What are the possible six digit combinations you make out of the numbers 1 6?
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∙ 11y ago
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David Macejkovic
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If you wanted to know how many 5 digit combinations you could make using only 2 digits how would you solve this?
Simple enough to solve. The answer is a power of two. Assuming you have two possible digits, say for example, 3 and 4, then you simply have to multiply it by how many numbers you want to get the total number of combinations. Each number can be 3 or 4 in this case, and you have 5 numbers. That's two to the fifth. Five combinations of any two numbers. 2x2x2x2x2. The answer is 32 combinations.
How many 3 digit numbers can be formed?
It depends upon whether the numbers can be used more than once. If the numbers can be used more than once, then there are 1,000 possible combinations; if not, then there are 720 possible combinations. ========== Assuming you are talking about integers you can calculate it this way: you can have any one of 9 digits for the first digit (zero is excluded because that would make it only a 2 digit number) You can have any one of 10 digits for the second and any one of 10 digits for the third digit. That gives you 9x10x10 = 900 different combinations for 3 digit numbers (not 1000). If you can include both negative and positive numbers as different numbers you get twice as many (2x900=18000). If you can count decimal numbers as 3 digit numbers (i.e. not restricted to integers) you can have 900 integers, 900 with the form xx.x, 1000 with the form x.xx (if zero is permitted to be the first digit and count as one of the 3 digits) or 900 (if a leading zero is NOT counted as one of the 3 digits). If a leading zero is NOT counted as one of the 3 digits, you could also have 1000 numbers of the form 0.xxx or just .xxx
What 4 digit numbers can you make out of 0123?
what is the greatest 4-digit number you can make with the digit 0,1,2,3, and 4
How many 4 digit numbers can you make with 3358?
12 numbers
How many 4 digit combinations can be made with numbers 3 7 9 4?
The number of combinations possible for taking a specified sub-set of numbers, r, from a set, n, isC(n,r) = n!/[r!(n-r)!]In this case, n = 4 and r = 1, soC(n,r) = 4!/1!3!C(n,r) = 24/6C(n,r) = 4Therefore, there are four possible combinations of the numbers 3, 7, 9 and 4.==================================This contributor strongly disagrees, but is leaving the original answer hereso that others can shop and compare.With their commas, parentheses, and factorials, the formulas are certainly impressive.Only the conclusion is wrong.The question specified "4-digit" combinations, so 'n' and 'r' are both 4.Now, to come down off the pedestal and make it understandable as well asformally rigorous ..."Combination" really means different groups of 4 digits that you can select outof the four you gave us. There's only one of those groups.If you actually want to know how many different 4-digit numbers you can makefrom them, then what you want is called "permutations" of the four digits, andyou can think of it this way, without using 'n', 'r', or parentheses or factorials:The first digit can be any one of 4. For each of those . . .The second digit can be any one of the remaining 3. For each of those . . .The third digit can be either one of the remaining 2.So the total number of different ways to line them up is (4 x 3 x 2) = 24 different 4-digit numbers.
How many possible combinations can you make out of a 5 digit alpha numeric numbers?
90
What are the possable 3 digit combinations you can make out of 1-9?
The number of possible 3 digit combinations you can make out of 1-9 with outrepeated digits is:9C3 = 9!/(3!(9-3)!) = 84
How many five-digit numbers can you make with 0 1 2?
Not sure what a "didget" is. It is possible to make 18 5-digit numbers.
What are the possible 4 digit combinations you make out of the numbers 1-6 Same numbers in each row 1111 2222 3333 4444 5555 6666?
You can make 24 4-digit numbers. 1111 2222 3333 4444 5555 6666 1234 2345 3456 1356 6543 5432 4321 1432 ok i give up!
How many 4 digit password combinations can you make with the numbers 4 5 6 9 and 0?
There are 5*4*3*2 = 120 such numbers.
What prime numbers make 50?
There are many combinations possible The most popular combination is 2,3,5,17,23.
What are all the possible number combinations to make 4 digit even numbers for 7 5 4 and 8?
4578, 4758, 5478, 5748, 5874, 5784, 7584, 7854, 7458, 7548, 8754, 8574
If you wanted to know how many 5 digit combinations you could make using only 2 digits how would you solve this?
Simple enough to solve. The answer is a power of two. Assuming you have two possible digits, say for example, 3 and 4, then you simply have to multiply it by how many numbers you want to get the total number of combinations. Each number can be 3 or 4 in this case, and you have 5 numbers. That's two to the fifth. Five combinations of any two numbers. 2x2x2x2x2. The answer is 32 combinations.
How many combinations can a six digit numeric code make from 0 to 9 and numbers can be used more than once?
106 or a million.
How many 4 digit combinations can be make with the numbers 3 7 1 6?
Just one. In a combination, the order of the digits does not matter.
How many three digit numbers can you make using the numbers 2 3 5 8 and 9?
There should be 125 combinations, but if you can only use the numbers once for each 3 digit number, then it should be 120. It's 125 because if the digits could be used more than once for each number then its the number on possible numbers (so there's 5 in this case, 2, 3, 5, 8, and 9) to the power of how big the digit is (in this case it's 3) It's 120 because if the digits could only be used once in each number then the combinations would be the number of possible numbers (again, 5) factorial, so 5!
How many 3 digit numbers can be formed?
It depends upon whether the numbers can be used more than once. If the numbers can be used more than once, then there are 1,000 possible combinations; if not, then there are 720 possible combinations. ========== Assuming you are talking about integers you can calculate it this way: you can have any one of 9 digits for the first digit (zero is excluded because that would make it only a 2 digit number) You can have any one of 10 digits for the second and any one of 10 digits for the third digit. That gives you 9x10x10 = 900 different combinations for 3 digit numbers (not 1000). If you can include both negative and positive numbers as different numbers you get twice as many (2x900=18000). If you can count decimal numbers as 3 digit numbers (i.e. not restricted to integers) you can have 900 integers, 900 with the form xx.x, 1000 with the form x.xx (if zero is permitted to be the first digit and count as one of the 3 digits) or 900 (if a leading zero is NOT counted as one of the 3 digits). If a leading zero is NOT counted as one of the 3 digits, you could also have 1000 numbers of the form 0.xxx or just .xxx
