# Baryon operators for $N_f<N_c$

Simple question. The answer is probably obvious, but I just don't see it.

In section 3 of this paper http://arxiv.org/abs/hep-th/9509066 it is stated that in SUSY QCD if $N_f<N_c$ the classical moduli space space is in terms of mesons $M^i_j=Q^i\bar{Q_j}$, but that for $N_f \geq N_c$ we can also have baryons $B^{i_1...i_{N_c}}=Q^{i_1}...Q^{i_{N_c}}$. Can someone please explain this to me? Why can't I have baryons in the first case? I thought I can make a proton out of up and down quarks with $N_f=2, N_c=3$ ...

Ref (page 5)

From the quark operators $\Phi_{ci}, \bar \Phi^{ci}$, where $c$ is a colour indice, and $i$ is a flavour indice, one may define :

Mesons operators : $M^j_i = \bar \Phi^{jc}\Phi_{ci}$

Baryons operators : $B_{i_1...i_n} = \epsilon^{c_1c_2...c_{N_c}}\Phi_{c_1i_1}\Phi_{c_2i_2}...\Phi_{c_{N_c}i_{N_c}}$

$\quad \quad \quad \quad \quad \quad \quad \bar B^{i_1...i_n} = \epsilon_{c_1c_2...c_{N_c}}\bar \Phi^{c_1i_1}\bar \Phi^{c_2i_2}...\bar \Phi^{c_{N_c}i_{N_c}}$

Due to the antisymmetry of the $\epsilon$ symbols, the non-null baryons operators correspond to a set of distinct $i_1...i_n$. Counting the number of baryon operators $B_{i_1...i_n}$ (for instance) is the same thing as counting the number of possible different $N_c$ indices $i_1, i_2...i_{N_c}$ among $N_f$ possible indices.

So the number of baryons operators $B_{i_1...i_n}$ is $\begin{pmatrix} N_f\\N_c\end{pmatrix}$. So for $N_f < N_c$, there are not baryons operators.