= Solution
{c}
The <stoichiometry> fixes the site <probabilities> to $1/4$ <gold> and $3/4$ <copper>. Every site in the average <crystal structure> therefore has the same <atomic scattering factor>
$$
\bar f=\frac14f_{\rm Au}+\frac34f_{\rm Cu}.
$$
The average <structure factor> is
$$
\bar S_{hkl}=\bar f\left[1+(-1)^{k+l}+(-1)^{h+l}+(-1)^{h+k}\right].
$$
At $(110)$ the bracket vanishes, while at $(111)$ it is four. Hence
$$
\boxed{I_{110}^{\rm Bragg}=0,\qquad
\bar S_{111}=4\bar f=f_{\rm Au}+3f_{\rm Cu}=S_{111}^{\rm ordered}.}
$$
The $(111)$ <powder diffraction multiplicity> is unchanged, so its coherent <Bragg diffraction> <wave intensity> equals the ordered value under the stated equal-<lattice constant> approximation.
The zero refers to the coherent <Bragg diffraction> peak of the average <crystal structure>. A random <alloy> can still give <occupational diffuse scattering> at the same <scattering vector>: the average of the squared amplitude includes fluctuations as well as the square of the average amplitude. Such background is not a surviving $(110)$ <Bragg diffraction> peak.
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