For 2-pentyl-X with a strong base under E2 conditions, which is the correct order of increasing proportion of the Hofmann product, pent-1-ene, for X = $\mathrm{Br}$, $\mathrm{F}$ and $-\mathrm{N}{}^{+}(\mathrm{CH}{}_3){}_3$?

The poorer the leaving group, the later the C-X cleavage and the more carbanion-like the E2 transition state, which favours removal of the least hindered, most acidic hydrogen. Bromide is a good leaving group and gives mostly Saytzeff product. Fluoride is poor and gives mostly Hofmann product, and the positively charged $-\mathrm{N}(\mathrm{CH}{}_3){}_3{}^{+}$ group, which is bulky and also strongly acidifies neighbouring hydrogens, gives nearly exclusive Hofmann product.