I can't seem to understand.
I thought about it this way. We take a first example:
$222aaa$ for each blank space we have $5$ positions, so in this case the answer would be $125$ ($5 \times 5 \times 5$) numbers.
Now the position of the twos can change so we calculate the numbers for that taking this example:
$222aaa$ and the possibilities for this are the number of ways you can arrange $6$ digits ($6!$) and divide by repetitions so divided by $2 \cdot (3!)$ so answer $= 20$
So final answer should be $125 \cdot 20 = 2500$. But this answer is wrong and I don't understand why.