The four numbers that can be used in combinations or differences to make all numbers from 1 to 30 are 1, 2, 4, and 8. These numbers are powers of 2, which allows for the creation of all numbers from 1 to 30 through various combinations and differences. By using these four numbers strategically, one can generate any integer between 1 and 30.
There are 10,000 possible combinations, if each number can be used more than once.
If each number can be used only once then there are just four: 2345, 2346, 2456 and 3456. There are no others because the order of the digits does not matter in combinations.
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!
The answer depends on what mathematical operations are permitted and on whether or not all four number must be used.
Well, honey, if you're looking for 7-number combinations out of 10 numbers, that's just basic math. You take 10 choose 7, which is 120. So, there you have it, 120 possible combinations. Math doesn't have to be complicated, darling.
There are no differences in the numbers.
106 or a million.
WRONG ANSWER NB !!If repeated numbers are permitted the answer is 8x8x8x8x8=32768 (85)If the numbers can only be used once in each combination, the answer is 8x7x6x5x4= 6720, where ABCDE has [n!/ (n-r)!] or [8! / 3!] combinations for 8 numbers used 5 at a time.NONONONO THIS IS NOT THE RIGHT FORMULA IT SHD BE [n ! / (n-r)! x r! ] = ONLY 56 COMBINATIONS where each number can only be used once.8x7x6x5x4x3x2x1 / (3x2x1) ( 5x4x3x2x1) = 56from napoleon solo
NOT AN ANSWER:More acceptable combinations:1231 3213 2321More unacceptable combinations:1112 2222 3113Thanks.
If all numbers can be used as many times as wanted then there are 109 = one billion combinations. If each number can be used only once, there are 10!/(10-9)! = 10!/1! = 10! = 3628800 combinations. * * * * * Clearly answered by someone who does not know the difference between PERMUTATIONS and COMBINATIONS. The combination 123456789 is the same as the combination 213456789 etc. All in all, therefore, there are only ten combinations which use each digit at most once.
If every number can be used as many times as you like, there are 104 = 10000 different combinations. If each number can only be used once, there are 9!/(9 - 4)! = 5040 combinations.
There are 10,000 possible combinations, if each number can be used more than once.
Assuming each number can be used more than once.... 2401 possible combinations.
Oh, dude, you're making me do math now? Alright, so if you have three numbers and you're asking how many combinations you can make with those three numbers, it's like a little math puzzle. Each number can be used multiple times, so it's like a little party for those numbers. The total number of combinations you can make with three numbers is 27. That's like having 27 different outfits to choose from for a night out, but with numbers.
Whoever asked this question you spelled combinations wrong and combinations are what are used to see how many pairs or groups you can make out of objects.
There are 36 possible characters (26 letters + 10 numbers) that can be used in each position of the 11-digit combination. Therefore, the total number of possible combinations is 36^11, which is approximately 7.52 x 10^17. This means there are over 750 quadrillion possible 11-digit combinations of letters A-Z and numbers 0-9 when combined.
If the same 7 digits are used for all the combinations then n! = 7! = 7*6*5*4*3*2*1 = 5040 combinations There are 9,999,999-1,000,000+1=9,000,000 7-digit numbers.