Two ways. Sum of all weights is $45$. Hence, the three weights $w_1<w_2<w_3$ on RHS should weigh (together) at least $23$. Moreover, $w_1+w_2+w_3\le 24$ (by taking $9,8,7$). There is a unique way to make it $24$. For $23$; if $w_3<9$, then $w_1+w_2+w_3\le 21$. Hence $w_3=9$ and $w_1+w_2=14$. From here, $w_2=8$ and $w_1=6$ is the only possibility. Thus, there is a total of two different such ways.