As there are 4 numbers and they are all different, they can be arranged in:
4 choices for first number, leaving
3 choices for the second number for each choice of the first, leaving
2 choices for the third number for each of the choices of the first two, leaving
1 choice for the last number for each of the choices for the first three
giving:
number of choices = 4 x 3 x 2 x 1 = 24 possible orders in total.
-2
Assuming leading zeros are not permitted, then: If repeats are not allowed there are 30 possible numbers. If repeats are allowed there are 60 possible numbers.
If the numbers can be repeated and the numbers are 0-9 then there are 1000 different combinations.
450
9!/6!, if the six different orders of any 3 digits are considered distinct combinations.
5 293 and 2 935 and 3 532 orders of numbers from least to greatest
4 of them. In a combination the order of the numbers does not matter.
-2
3, 19 and 43 is one of many possible answers.3, 19 and 43 is one of many possible answers.3, 19 and 43 is one of many possible answers.3, 19 and 43 is one of many possible answers.
Assuming leading zeros are not permitted, then: If repeats are not allowed there are 30 possible numbers. If repeats are allowed there are 60 possible numbers.
If the numbers can be repeated and the numbers are 0-9 then there are 1000 different combinations.
1004
450
there are 22
9!/6!, if the six different orders of any 3 digits are considered distinct combinations.
There are infinitely many possible answers.
0.30000001 is one of infinitely many possible numbers.