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This is a permutation problem.
The digits are: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
The 1st digit has 9 possibilities (it cannot be zero).
Thus, there are 9(9P9) = 9(9!) = 9(362,880) = 3,265,920 such numbers, because:
The 2nddigit has 9 possibilities (because we are left with 9 digits).
The 3rd digit has 8 possibilities.
The 4th digit has 7 possibilities.
The 5thdigit has 6 possibilities.
The 6thdigit has 5 possibilities.
The 7thdigit has 4 possibilities.
The 8thdigit has 3 possibilities.
The 9thdigit has 2 possibilities.
The 10thdigit has 1 possibility.
so that, 9*9*8*7*6*5*4*3*2*1= 3,265,920.
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