To have product 8 the 10 digits have to be one of these types:
A. those with this group of digits: 2,2,2,1,1,1,1,1,1,1
B. those with this group of digits with 2 left of 4: 2,4,1,1,1,1,1,1,1,1
C. those with this group of digits with 4 left of 2: 4,2,1,1,1,1,1,1,1,1
D. those with this group of digits: 8,1,1,1,1,1,1,1,1,1
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There are 10 positions to choose in a 10-digit number
We calculate the number of 10-digit numbers of type A:
The 3 2's can be placed in 10C3 or 120 positions
We calculate the number of 10-digit numbers of type B
The 2 and 4 with 2 left of 4 can be placed in 10C2 or 45 positions
We calculate the number of 10-digit numbers of type C
The 4 and 2 with 4 left of 2 can be placed in 10C2 or 45 positions
We calculate the number of 10-digit numbers of type D
The 8 can be placed in 10C1 or 10 positions
That's 120+45+45+10 = 220 ways
Edwin