More then 30 how do i know that because i did it it took me more then 3 hours
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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 900,000,000 nine-digit numbers that do not have leading zeros, or the 999,999,999 nine-digit number including leading zeros.
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.
There are a lot of them, so I'm sure you don't want us to list them all. If you want 0-9, and are happy with repeat digits, then it's 10*10*10*10 = 10000 combinations. If it's 1-9 and are happy with repeat digits, then it's 9*9*9*9 = 6561 combinations. 0-9 without repeat digits is 10*9*8*7 = 5040 combinations 1-9 without repeat digits is 9*8*7*6= 3024 combinations
If you are not allowed to repeat the digits then the answer is clearly 0.If you are allowed to repeat digits then the only way you can possibly reach a product of 4 is by using a combination of 3 "1's" and 2 "2's". The possible combinations are therefore:22111212112112121112122111212112112112211121211122Thus there are 10 numbers which meet the criteria.
13,579 or .13579 . If I'm allowed to repeat digits, then you are 11,111 or .11111 .
repetition
If the digits can repeat, then there are 256 possible combinations. If they can't repeat, then there are 24 possibilities.
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.
Assuming you can repeat digits (like the number 1228 for example), there are 84 = 4096.If you can't repeat digits then it is equivalent to 8!/4! = 1680.
The first digit can be one of five integers (1 to 5) The second - fourth digits one of six integers (0 to 5). So: number of (valid) combinations is 5*6*6*6=1080 * * * * * This gives the number of permutations, not combinations. I am working on the latter.
If the numbers are allowed to repeat, then there are six to the fourth power possible combinations, or 1296. If they are not allowed to repeat then there are only 360 combinations.
There are a lot of them, so I'm sure you don't want us to list them all. If you want 0-9, and are happy with repeat digits, then it's 10*10*10*10 = 10000 combinations. If it's 1-9 and are happy with repeat digits, then it's 9*9*9*9 = 6561 combinations. 0-9 without repeat digits is 10*9*8*7 = 5040 combinations 1-9 without repeat digits is 9*8*7*6= 3024 combinations
0000-9999 (10x10x10x10 or 104) = 10,000 possible combinations allowing for repeated digits. If you are not able to repeat digits then it's 10 x 9 x 8 x 7 or 5,040 possible combinations without repeated digits.
If you are not allowed to repeat the digits then the answer is clearly 0.If you are allowed to repeat digits then the only way you can possibly reach a product of 4 is by using a combination of 3 "1's" and 2 "2's". The possible combinations are therefore:22111212112112121112122111212112112112211121211122Thus there are 10 numbers which meet the criteria.
That depends on . . . -- how many digits you want in each combination -- whether it's allowed to repeat the same digit -- whether the order of the digits in the combination matters, i.e. whether you actually want combinations or permutations. Each choice has a different answer.
13,579 or .13579 . If I'm allowed to repeat digits, then you are 11,111 or .11111 .
repetition
-- If you're allowed to repeat digits in one combination,then there are 104 = 10,000 possibilities.-- If all four of the digits must be different, then there are(10 x 9 x 8 x 7) = 5,040 possibilities.