A seven-digit number made up of all distinct digits 8, 7, 6, 4, 2, x, y is divisible by 3. The possible number of ordered pairs (x, y) is:

For a number to be divisible by 3, the sum of its digits must be divisible by 3. In this case, the digits are 8, 7, 6, 4, 2, x, y. The sum of these digits is:

$8+7+6+4+2+\mathrm{x}+\mathrm{y}$

Now, the sum needs to be divisible by 3. Let's consider the given digits:

$27+\mathrm{x}+\mathrm{y}$

For this expression to be divisible by 3, the sum $\mathrm{x}+\mathrm{y}$ must be a multiple of 3. The possible pairs (x, y) that satisfy this condition are:

(0, 0) (1, 2) (2, 1) (2, 4) (4, 2) (4, 5) (5, 4) (7, 5)

So, there are indeed eight ordered pairs (x, y) that make the seven-digit number divisible by 3.