Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2007 / B / 7 / c / Solution

For type $(1^n)$, all labels are distinct, so a semistandard tableau is precisely a standard Young tableau. Thus $K_{\lambda ,(1^n)}=f^{\lambda }$, and Young's rule givesThis is also the decomposition of the regular representation, since tabloids with one entry per row are naturally indexed by all permutations.