= Solution
For <quarks> in an $N_c$-dimensional internal space $V$, complete antisymmetrization of three labels corresponds to a <Young diagram> with one column of three boxes, the partition $(1,1,1)$. Its <representation> is the <exterior power> $\bigwedge^3V$ and has <dimension>
$$
\dim\bigwedge^3V=\binom{N_c}{3}.
$$
For $N_c=2$, the column is too tall: $\bigwedge^3\mathbb C^2=0$. With only two labels at least two of the three coincide, and the <Young tableau> column antisymmetrizer annihilates the state. For $N_c=3$, the filled column $r,g,b$ spans a one-dimensional determinant <representation>. It is a <colour singlet> of <SU(3)> because every group element has determinant one.
Equivalently, the <tensor product of group representations> decomposes as
$$
\boxed{3\otimes3\otimes3=10_{[3]}\oplus8_{[2,1]}\oplus8_{[2,1]}\oplus1_{[1,1,1]}.}
$$
The two mixed-symmetry terms account for $16$ dimensions, so the total is $10+8+8+1=27$. Antisymmetry alone establishes \b[at least] three internal states. Exactly three follows in the usual <special unitary group> model when the observed three-<quark> <baryon> must be a <colour singlet>: for $N_c>3$ the antisymmetric third <exterior power> has dimension greater than one and is not a singlet. This extra assumption should be made explicit.
Back to article page
