To find $\dim(D^{\mu})=\dim\frac{S^{\mu}}{S^{\mu}\cap S^{\mu\perp}}$

Question

Let $S_6$ be the symmetric group of degree $6$ and $F$ be any finite field of characteristic $2.$ Then $2$-regular partition of $6$ are $(5,1)$, $(4,2)$ and $(3,2,1)$ . I have to find $$\dim(D^{\mu})=\dim\frac{S^{\mu}}{S^{\mu}\cap S^{\mu\perp}}$$where $S^{\mu}$ is the Specht module corresponding to partition $\mu.$ I know $$\dim(D^{(5,1)})=\dim\frac{S^{(5,1)}}{S^{(5,1)}\cap S^{(5,1)\perp}}=4$$ as $\dim S^{(n-1, 1)}=n-1$ and $S^{(n-1, 1)\perp}\subseteq S^{(n-1, 1)}$ as $n=6.$ For more details see at page number $18$ of representation theory of symmetric groups by G.D.James book. But I am unable to find dimensions of $D^{(4,2)} $and $D^{(3,2,1)}.$ I think to find rank of Gram matrix is very lengthy process. Please help me to obtain these dimensions. Thanks.