Given the involved numbers are relatively small, this problem can be solved using casework:

$\underline{no\ yellow\ ball}$

• If no white ball chosen, then all the three balls are red. There is only $1$ way.
• If $1$ white ball chosen, then we have $1$ white and $2$ red. There are $3$ ways.
• If $2$ white balls chosen, then we have $2$ white and $1$ red, There are $3$ ways.

$\underline{1\ yellow\ ball}$

• If no white ball chosen, then we have $1$ yellow and $2$ red. There are $3$ ways.
• If $1$ white ball chosen, the we have one color each. There are $3!=6$ ways.
• If $2$ white balls chosen, then we have $1$ yellow and $2$ white. There are $3$ ways.

Hence, the final answer is $1+3+3+3+6+3=\boxed{19}$ ways.