16 if you include 0 starting
12 if not
That is not correct.
0123 is a 4-digit number = (n)
therefore n! = 4! = 1*2*3*4 = 24 (if 0 is included as I assume that is the question)
123 is a 3-digit number = (n)
Therefore n! = 3!= 1*2*3 = 6
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With four digits (0, 1, 2, 3) and no repetition allowed, the number of possible combinations can be calculated using the permutation formula. Since the order matters, the number of permutations is 4! (4 factorial) which equals 4 x 3 x 2 x 1 = 24. Therefore, there are 24 possible number combinations with the digits 0, 1, 2, and 3.
Oh, what a happy little question! With those four numbers, there are 24 possible combinations. Just think of all the beautiful possibilities you can create with those numbers, like painting a serene landscape with each combination. Remember, there are no mistakes, just happy little accidents in math and art.
Just 4: 123, 124, 134 and 234. The order of the numbers does not matter with combinations. If it does, then they are permutations, not combinations.
As there are 26 letters in the alphabet. You can calculate the number of combinations by multiplying 26x26x26, giving you the answer 17576.
if i got everything right, its 1000
4 of them. In a combination the order of the numbers does not matter.
It depends on whether the order of the numbers is important or not. For example, if 123456 is seen as a different code to 213456 then there are many more possible solutions.If the order is unimportant, the number of possible combinations is equal tobinomial coefficient(15,6) = 5005If the order is important, then the number of possible permutations is equal to15! = ~1.3x1012