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Using only sums and differences, and not necessarily all four numbers, 1, 3, 9 and 27 will make all numbers from 0 to 40.

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Q: What 4 numbers can be used in combinations or differences to make all numbers from 1 to 30?
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What are all the possibilities of 4-digit numbers from numbers 0-9?

There are 10,000 possible combinations, if each number can be used more than once.

How many 4 number combinations are their using numbers 2-6?

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.

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 combinations will EQUAL 24 from the numbers 5 3 3 2?

The answer depends on what mathematical operations are permitted and on whether or not all four number must be used.

How many 7 number combinations in 10 numbers?

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)

Related questions

What are two major differences between numbers used in math and numbers used doing physics?

There are no differences in the numbers.

What four numbers times to make 90?

Lots of different combinations can be used, one being: 1x2x3x15 We can see that in any order we multiply these numbers in it will equate to 90.

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 5 digit combinations can you make with 8 numbers?

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

How many different 4 digit combinations are there for numbers 1 2 3 (assuming all three numbers are used and no numbers can be repeated more than once) (For example, 1233 is acceptable, but 1133 is not.)?

NOT AN ANSWER:More acceptable combinations:1231 3213 2321More unacceptable combinations:1112 2222 3113Thanks.

How many combinations of 9 digit numbers can be produced from the numbers 0 through 9?

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.

How many 4 digit combinations can be made from the numbers 0 to 9?

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.

What are all the possibilities of 4-digit numbers from numbers 0-9?

There are 10,000 possible combinations, if each number can be used more than once.

How many combination of 4 numbers are there in the numbers 1 to 7?

Assuming each number can be used more than once.... 2401 possible combinations.

What are combinateions in math?

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.

How many possible combinations of a 7-digit number?

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.

How many combinations of 5 numbers are there in 55 numbers?

combinations of 55If there is not repetition in numbers used......55 x 54 x 53 x 52 x 51 * * * * *No. That is the number of permutations. For combinations the order of the number does not matter so that 1,2,3,4,5 is the same as 1,3,2,4,5 or 1,4,2,3,5 etc.There are 5*4*3*2*1 = 120 such orderings for every set of 5 numbers. So the correct answer is55*54*53*52*51 / 120 = 3478761