Which pair of reagents would give cetyltrimethylammonium bromide, $[\mathrm{C}{}_{16}\mathrm{H}{}_{33}\mathrm{N}(\mathrm{CH}{}_3){}_3]{}^{+}\mathrm{Br}{}^{-}$, directly in one step?

Trimethylamine is a nucleophile that attacks the primary alkyl bromide in an ${\mathrm{S}}_{\mathrm{N}}2$ reaction: $(\mathrm{CH}{}_3){}_3\mathrm{N}+\mathrm{C}{}_{16}\mathrm{H}{}_{33}\mathrm{Br}\to [\mathrm{C}{}_{16}\mathrm{H}{}_{33}\mathrm{N}(\mathrm{CH}{}_3){}_3]{}^{+}\mathrm{Br}{}^{-}$. One equivalent of $\mathrm{CH}{}_3\mathrm{Br}$ with a primary amine gives a mixture of lower alkylation products, and alcohols or acids do not alkylate the amine.