To take out 7 numbers whose sum is a multiple of 3, it is sufficient to make the sum of the remaining two numbers also a multiple of 3. Among these 9 numbers, there are 3 numbers each that leave a remainder of $0, 1, 2$ when divided by 3. To take out 2 numbers whose sum is a multiple of 3, it can be $0+0$, or $1+2, 0+0$ has 3 ways of selection, $1+2$ has $3 \times 3=9$ ways of selection. In total, there are 12 ways of selection.
15. From 11, 22, 33, 44, 55, 66, 77, 88, 99, select 7 numbers such that their sum is a multiple of 3, there are ( ) different ways.