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If using numbers 1 thru 49 how many 6 digit combinations can be made?
Using the formula n!/r!(n-r)! where n is the number of possible numbers and r is the number of numbers chosen, there are 13983816 combinations of six numbers between 1 and 49 inclusive.
How many combinations in sets of 5 using the numbers 1 thru 10?
There are 252 combinations.
What is the product of the 10 single digit numbers?
362,880 edit: 3,628,800 edit: Sorry, Whizkid. You included '10', whereas the question clearly stated "single-digit numbers". Now that we think about it, the "10 single digit numbers" must include zero, so the product is zero. But for the digits 1 thru 9, the product is still 362,880 .
How many 3 digit numbers have at least one repeated digit?
There are a total of 900 3-digt numbers (100 thru 999); Of these there are 9x9x8 = 648 numbers that do not contain repeating digits (first position: 9 possibilities, second place 9 possibilities; the 10 digits minus the one used in the first place. and third place 8 possibilities; 10 digits minus the two used in the first and second places) So, there should be 900-648=252 3-digit numbers with at least one repeating digit.
How many possible 4 digit 0 thru 9 codes?
If the first digit can be zero, there are 5,040. If the first digit can't be zero, there are only 4,536.
If using numbers 1 thru 49 how many 6 digit combinations can be made?
Using the formula n!/r!(n-r)! where n is the number of possible numbers and r is the number of numbers chosen, there are 13983816 combinations of six numbers between 1 and 49 inclusive.
How many combinations in sets of 5 using the numbers 1 thru 10?
There are 252 combinations.
Sum of the digits of a number equals the number.Generate such numbers?
It appears that only single digit numbers work (0 thru 9)
How many 5-number combinations are there with the numbers 1 through 35 please list them all?
how mant 5-number combinations are there with tnumbers 1 thru 35, please list them all?
What is the product of the 10 single digit numbers?
362,880 edit: 3,628,800 edit: Sorry, Whizkid. You included '10', whereas the question clearly stated "single-digit numbers". Now that we think about it, the "10 single digit numbers" must include zero, so the product is zero. But for the digits 1 thru 9, the product is still 362,880 .
How many 5 digit numbers are there with distinct numbers?
I suspect you mean "without repeated digits", and I'll answer it that way.Here's how I would construct all the 5-digit numbers without repeated digits:The first digit can be any one of 9 (1 thru 9 but not zero). For each of these . . .The second digit can be any one of 9 (zero thru 9 but not the same as the first one). For each of these . . .The third digit can be any one of the remaining 8. For each of these . . .The fourth digit can be any one of the remaining 7. For each of these . . .The fifth digit can be any one of the remaining 6.Total number of possibilities = (9 x 9 x 8 x 7 x 6) = 27,216
How many different 3 digit combinations can be made using 0 thru 9 and they can repeat?
Using the word "combinations" in the English sense (as opposed to mathematical sense the expert has used) where it often used in the mathematical sense of the word "permutations": Assuming the hundreds digit must be at least 1 (eg 99 = 099 is not considered a three digit number), then: 9 x 10 x 10 = 900.
How many different four digit combinations are there of the numbers zero through nine without using the same number twice in each combination?
The first digit can be any one of the ten symbols, 0 thru 9. For each of those . . .The second digit can be any one of the 9 that weren't used in the first place. For each of those . . .The third digit can be any one of the 8 that you haven't used yet. For each of those . . .The fourth digit can be any one of the 7 that haven't been used yet.Total number of possible arrangements = (10 x 9 x 8 x 7) = 5,040
How many 3 digit numbers have at least one repeated digit?
There are a total of 900 3-digt numbers (100 thru 999); Of these there are 9x9x8 = 648 numbers that do not contain repeating digits (first position: 9 possibilities, second place 9 possibilities; the 10 digits minus the one used in the first place. and third place 8 possibilities; 10 digits minus the two used in the first and second places) So, there should be 900-648=252 3-digit numbers with at least one repeating digit.
How many possible 4 digit 0 thru 9 codes?
If the first digit can be zero, there are 5,040. If the first digit can't be zero, there are only 4,536.
What is round to the nearest 10 on31?
you round up on a number if the last digit is 5 thru 9 you round down if the last digit is 4 thru 1 therefore 31 rounded to the nearest ten is 30
How many 4 number combinations are there in the numbers 0 to 9?
The easy answer is 0000 thru 9999, which is 10,000 combinations. A way to calculate this is by Nxwhere N is the number of digits possible (here, 10 different digits arepossible) 0 through 9, and x is the number of places available (here we have 4 places available for a 4 digit number)note the formula is N raised to the x powermeaning, 104or 10 X 10 X 10 X 10 = 10,000 combinationsAnother exampleHow many combinations are available using digits 0-3 for a 2 digit number? The possibilities are: 00 01 02 03 10 11 12 13 20 21 22 23 3031 32 33or 16 possibilitiesthe formula would tells us:Nx is 42 or 4 x 4 = 16* * * * *THE ABOVE ANSWER IS WRONG!Interesting but completely wrong since the answer confuses permutations and combinations. The combination 12 is the same as the combination 21.The number of 4 digit combinations from the numbers 0 to 9 is 10C4 = 10!/[4!*6!]=10*9*8*7/(4*3*2*1) = 210.
