Because of the question's ambiguity, it is hard to tell exactly what is meant. However, to cover all possibilities:
If there are four numbers and each is used once, there are 4! = 4 x 3 x 2 x 1 = 24 different combinations.
If there are the 10 digits used to make a four-digit number, using each once, we have 10!/(10-4)! = 10!/6! = 10 x 9 x 8 x 7 = 5040 different combinations.
Or does he mean this?
Using the binary numbering system.
Binary Decimal
0000 0
0001 1
0010 2
0011 3
0100 4
0101 5
0110 6
0111 7
1000 8
1001 9
1010 10
1011 11
1100 12
1101 13
1110 14
1111 15
750
256
None, because you do not have three even digits. None, because you do not have three even digits. None, because you do not have three even digits. None, because you do not have three even digits.
234, 243, 324, 342, 423, and 432.
9
The final digit must be a 2 to form an even number. The first digit may be any of three remaining digits (1, 3, 5) while the second digit may be any of the two remaining digits. All together, that makes 3*2 = 6 distinct even numbers.
If you want 4-digit numbers, there are 24 of them.
If you use each number once, there are six combinations.
Assuming each of the given digits can be used only once, the answer is 24. If not, the answer is infinity.
2
6
There are six choices, the 5 digits and the one you leave blank. That makes 6! or 720 different numbers.
Using the combination C(4,3) I got the answer 24. Its either a combination or a permutation.
There are 5*4*3 = 60 such numbers.
19 do.
If the digits can be used more than once, then 900. If not, then 648.
There are 9*9*8*7 = 4536 such numbers.
24