8. How many 10-digit positive integers abcdefghij are there, with all digits being different and satisfying the conditions $a+j=b+i=c+h=d+g=e+f=9$?

Note: a number cannot start with 0.
(A) 3456
(B) 3528
(C) 3645
(D) 3840
(E) 5040.
8. The answer is (A). Let's call, as in the text, abcdefghij the 10 digits of the number.

For numbers of the required form, fixing the first 5 digits $a, b, c, d, e$ uniquely determines the entire number due to the imposed condition (since we can derive $f=9-e, g=9-d, \ldots$).

On the other hand, if two identical digits or two digits that sum to nine were to appear among the first five digits, we would have a number that does not meet the conditions, because on one hand, it is required that all digits be different, and on the other, if two digits with a sum of nine were to appear among the first five, they would also appear—albeit in reverse order—among the last five, while we want all of them to be different.

It is therefore sufficient to count the 5-digit numbers (the first 5), with $a, b, c, d, e$ all different and such that no pair sums to 9.

Forgetting for now that a number cannot start with 0, we see that $a$ can be chosen in 10 ways, $b$ in 8 ways (all digits except $a$, already used, and 9-a), $c$ in 6 ways (all digits are possible except $a, b$ and $9-a, 9-b), d$ in 4 ways, and $e$ in 2 possible ways.

From these, we must subtract the numbers that start with the digit zero, which are (given that $a=0$, at this point, is fixed) $8 \cdot 6 \cdot 4 \cdot 2$ for the same reasoning as above (8 choices for the digit $b, 6$ for $c, 4$ for $d$ and 2 for $e$).

The answer to the problem is therefore $10 \cdot 8 \cdot 6 \cdot 4 \cdot 2 - 8 \cdot 6 \cdot 4 \cdot 2 = (10-1) \cdot 8 \cdot 6 \cdot 4 \cdot 2 = 3456$.(Lombardo)