JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Permutations and Combinations

The letters A,E,F,M,R,R are arranged so the two R's are never adjacent. Words beginning with A use E,F,M,R,R: $\frac{5!}{2!}-4!=60-24=36$ valid words; those beginning with E give another $36$. Among words beginning with F, the second letter A comes first; with FA fixed and E,M,R,R to arrange, the blocks FAE$\dots$ and FAM$\dots$ give one valid word each. With FAR fixed, the remaining E,M,R fill the last three places without a leading R, giving in order FAREMR, FARERM, FARMER, FARMRE, so two words precede FARMER. The number of words before FARMER is $36+36+1+1+2=76$, placing FARMER at serial number $77$.