I would like to know how many 4digit combination of numbers in "0 to 9".?

Question

Example: 0123 can be combined in 24 different sets like 1230, 2301,3012, 1230,2301,3021 ect.
Is there any one can be helpful?
Is there any software to get all the combination?
Please help me.

Answer 1

There are 10 numbers to pick from for the first digit. There 9 remaining to pick from for the second -- since you can't pick the same one you picked for the first digit, one is eliminated. You're left with 8 different digits to pick for the third digit, and 7 for the fourth one.

So your answer for the number of four-digit numbers with no repeated digits is 10*9*8*7, which is 5,040.

In the general case, if you have D digits to pick from, and you are assembling a N-digit number (D >= N), the equation for the number of combinations with no repeated values is:

D! / (D-N)!

In this case it was
10! / (10-4)! =
10!/6! =
10*9*8*7*6*5*4*3*2*1 / (6*5*4*3*2*1) =
10*9*8*7

In the case of the example you provided (combinations of 0-1-2-3), you had four digits to choose from:
4!/(4-4)! =
4*3*2*1/1 =
24

Answer 2

If you can use the same number in diferent places and repeat any there are
10^4=10.000
If you don´t there are C(10,4) *P(4) = 10*9*8*7=5040

Answer 3

9999.

Any number you can write exists on your way to 9999,
so that is the number of numbers.

Answer 4

there are eactly 5040 ways/.