= Solution
Among the <representations> in <4/c>, the $\mathbf{10}$ of $SU(5)$ occurs in the $\mathbf{15}$ of $SU(6)$, while the simplest source of $\bar{\mathbf5}$ is $\bar{\mathbf6}$. Thus the economical <direct sum> containing the prescribed family is
$$
\boxed{R=\bar{\mathbf6}\oplus\mathbf{15}
\quad\downarrow SU(5)\quad
(\bar{\mathbf5}\oplus\mathbf{10})\oplus\mathbf5\oplus\mathbf1.}
$$
The extra $SU(5)$ <multiplets> are one $\mathbf5$ and one <trivial representation>. Calling $R$ a <representation> does not require it to be an <irreducible representation>; none of the five individual choices in <4/c> contains both required $SU(5)$ summands.
There is a further physical qualification if this is to be the entire chiral matter content of a consistent <grand unified theory>. Merely containing one <Standard Model> family does not guarantee cancellation of the cubic $SU(6)$ <gauge anomaly>. In the convention $A(\mathbf6)=1$, one has $A(\bar{\mathbf6})=-1$ and $A(\mathbf{15})=2$, so the displayed sum has $A(R)=1$. With these types of <representations>, the minimal anomaly-free completion is instead
$$
\boxed{R_{\mathrm{af}}=2\bar{\mathbf6}\oplus\mathbf{15}
\quad\downarrow SU(5)\quad
(\bar{\mathbf5}\oplus\mathbf{10})\oplus(\mathbf5\oplus\bar{\mathbf5})\oplus2\mathbf1.}
$$
Beyond one family this adds a <vectorlike gauge spectrum> $\mathbf5\oplus\bar{\mathbf5}$ and two <trivial representations>. The conjugate pair can receive a mass after the appropriate symmetry breaking, without removing the remaining family.<Number of fermion generations from a novel grand unified model by Pritibhajan Byakti and others>[Section 3, equations (14)–(16) and the following discussion: the SU(6) anomaly coefficients are -1 and 2, requiring twice as many conjugate fundamentals as antisymmetric rank-two tensors]{ref} The first boxed answer addresses representation containment; the second includes the extra consistency condition for a chiral <gauge theory>.
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