What are all of the possible 4 number combinations I can make with the didgits 1,2,3,4,5,6,7,8,9,0?

Answer 1

It depends on whether order matters and whether you can repeat digits.

If you can repeat digits and order matters then you have 10 choices for each of the four digits, so there are 10^4 = 10000 different numbers you can make.

If you cannot repeat digits and order matters then there are 10 choices for the first number, then 9, 8, and 7, so there are 10*9*8*7 = 5040 different numbers.

If however you cannot repeat digits and order does not matter (this is often called "combinations"), then there are 10!/(6!4!) = 210 sets of four digits.

Finally, if order does not matter, but you can repeat digits, there are are (10 + 4 - 1) choose 4 = 13*12*11*10/4*3*2*1 = 715 "multisets" of the digits 0-9 you can pick.

Answer 2

All the above are permutations...i.e. all the numbers from 0000 to 9999...

1, 2, 3 and (4-0)

1, 2, 4 and (5-0)

1, 3, 4 and (5-0)

1, 2, 5 and (6-0)

1, 3, 5 and (6-0)

1, 4, 5 and (6-0)

and so on...

Answer 3

If reusing numbers are not allowed, then this is a permutation (not a combination) problem, i.e. 1234 and 4123 are both answers. In that case, you'd have 10 * 9 * 8 * 7 = 5040

Answer 4

you can have any of those 10 digits as the first number in your 4 digit number. The same goes for all of the four digits in your number.

So you can have 10*10*10*10 different numbers = 10,000.

This is all the numbers from 0 to 9,999.

Answer 5

Clearly 10^4, everything from 0000 to 9999, right? Now if you don't want to repeat figures or stuff like this, it's an other story.

Answer 6

Depends on whether reuse is allowed, if yes, then the digits range from 0000 to 9999

If not, then combinatorials are used --> 10!/(6!4!)

Answer 7

10000 if you include 0000

Answer 8

0000
0001
0002
...
9999