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.
Without repeating any of the numbers (each used once) = 120 unique "sets" of 7 numbers (i.e. 1234567, 1234568, 1234569, 1234560, 1234578, 1234579, 1234570, 1234589, 1234580, 1234590) If the order matters, you have a much larger number of combinations. (see below) Each of the 120 sets can be arranged in 5040 ways. For a set of N numbers, the possible combinations using K numbers is N! / K! x (N-K)! 10! = 3,628,800 7! = 5040 (10-7)! = 3! = 6 The shortcut is (10 x 9 x 8) / (3 x 2) = 720/6 = 120 --- For rearrangement of numbers (ordered sets), there are 604,800 possible numbers N! / (N-K) ! = 10! / 3! = 3,628,800 / 6 = 604,800 --- For unlimited repetition, there are 10,000,000 (1 x 10^7)
There are no differences in the numbers.
106 or a million.
To make 24 by adding only, there are several combinations of numbers that can be used. For instance, you can use the numbers 1 through 23 in various combinations, such as 12 + 12, 10 + 10 + 4, or 8 + 8 + 8. Additionally, you can include multiple instances of smaller numbers, like 6 + 6 + 6 + 6. The total number of distinct combinations depends on the range of numbers you allow and any restrictions on repetitions.
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
To make 80 using numbers between 3 and 25, you can use addition, multiplication, or combinations of both. For example, the pairs (55, 25), (40, 40), and (30, 50) can be used in addition, while multiplication allows for combinations like (4 × 20) or (8 × 10). Additionally, you can find combinations of multiple numbers (like 20 + 20 + 20 + 20) to reach 80. However, the specific combinations will depend on how strictly you interpret the use of numbers.
To find how many ways to make 12 with three single-digit numbers (from 0 to 9), we can consider combinations of numbers that add up to 12. The valid combinations include repetitions of numbers, such as (3, 4, 5) and (2, 5, 5). However, since the order of addition matters, each unique combination can be arranged in multiple ways, which increases the total count. A systematic approach or generating functions can be used to enumerate all valid combinations and permutations, leading to a total of 29 distinct combinations.
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.
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 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.