Permutations

$9!*2 + 8!*6 + 7!*6 + 6!*2 + 5!*5 + 4!*1 + 3!*2 + 2!*1 + 1*1! = 999999$

This can be found easily by repeated division.

Now, make a table like(where right is the serial number)

$9 - 1$

$8 - 2$

$7 - 3$

$6 - 4$

$5 - 5$

$4 - 6$

$3 - 7$

$2 - 8$

$1 - 9$

$0 - 10$

And cut the number(in the serial number) succeeding the one you see in the multiple of the factorial in the above equation starting from left. For example-

As one can see that multiple of $9!$ is $2$ , we cut the number with serial number $3$ i.e. $7$ and so on.

Carefully note that after you cut any number, you have $1$ less than the previous total of numbers left, so serial numbers will change as well, like after first cut,

$9 - 1$

$8 - 2$

$6 - 3$

$5 - 4$

$4 - 5$

$3 - 6$

$2 - 7$

$1 - 8$

$0 - 9$

and so on.

The manner you cut the numbers will be something like - $2783915460$ and that's the desired number.

A programmatic solution using Python:

def genPermutation(A):
''' generate all permutations of a given array'''
    n = len(A)
    if n == 1:
        return [A]
    result = []
    for i in range(n):
        a = genPermutation(delete(A, i))
        for j in a:
            result += [[A[i]] + j]
    return result
def delete(A, n):
    '''delete value at index i of array A'''
    B = A.copy()
    L = len(B)
    assert(n < L)
    for i in range(n, L - 1):
        B[i] = B[i+1]
    return B[:-1]
arr = genPermutation([0,1,2,3,4,5,6,7,8,9])
print(arr[999999])

Begin with smallest number: 123456789, with 9 digits having:9! numbers. Next is 1023456789, with 1 at first having 9! numbers. Next is 2013456789. So the millionth number has 2 is the first digit. From 2013456789 to 2701345689 having 6*8! numbers, next to $2780134569: 6 \times (7!)$ next to $2783014569: 2 \times(6!)$ to $2783901456: 5 \times(5!)$ to $2783910456: 4!$ to $2783915046: 2 \times(3!)$ to 2783915460 is millionth number. $\boxed{2783915460}$