means the two boxes removed from the short column must remain stacked on top of each other. Thus, there are two possible irreducible representations $s$ that can be produced, implying we can subduce two possible irreducible representations of $S_9 \times S_3$, by removing these three boxes $$\young({\,}{\,},{\,}{\,},{\,},{\,},{\,},{\,},{\,})\, \young({\,},{\,},{\,})\qquad
\young({\,}{\,},{\,}{\,},{\,},{\,},{\,},{\,},{\,})\, \young({\,}{\,},{\,})$$ Thus, in total we’d get the following $S_9 \times S_3$ irreducible representations subduced from the $R$ given above (sum over all possible ways to remove boxes to get this result) $$\young({\,}{\,},{\,}{\,},{\,}{\,},{\,}{\,},{\,})\,\young({\,},{\,},{\,})   \oplus
\young({\,}{\,},{\,}{\,},{\,}{\,},{\,},{\,},{\,})\,\young({\,},{\,},{\,}) \oplus
\young({\,}{\,},{\,}{\,},{\,}{\,},{\,},{\,},{\,})\,\young({\,}{\,},{\,})  \oplus
\young({\,}{\,},{\,}{\,},{\,},{\,},{\,},{\,},{\,})\,\young({\,},{\,},{\,}) \oplus
\young({\,}{\,},{\,}{\,},{\,},{\,},{\,},{\,},{\,})\,\young({\,}{\,},{\,})  \oplus
\young({\,}{\,},{\,},{\,},{\,},{\,},{\,},{\,},{\,})\,\young({\,},{\,},{\,})$$ The dimensions of these representations are 42, 48, 96, 27, 54 and 8 respectively. This sums to give 275 which is indeed the dimension of $R$.

To construct the actual projector, we need only build an operator which will assemble the removed boxes in the correct way to produce $s$. We will give an example of how to construct this operator; the general case should be clear. Lets start with the representation $R$ shown, removing the boxes indicated $$\young({\,}{\,},{\,}{*},{\,},{*})$$ Our projector will act in the subspace spanned by the two sets of states $$|1\rangle =|\young({\,}{\,},{\,}{2},{\,},{1})\rangle\qquad
|2\rangle =|\young({\,}{\,},{\,}{1},{\,},{2})\rangle$$ We are using a Young-Yamonouchi basis. Thus, each state given above could 