All the arrangements, with or without meaning, of the word FARMER are written excluding any word that has two R appearing together. The arrangements are listed serially in the alphabetic order as in the English dictionary. Then the serial number of the word FARMER in this list is _________.

We begin by writing the six letters of the word FARMER in strict alphabetical order. In English dictionary order we have $$A \lt E \lt F \lt M \lt R.$$ Remember that the letter $$R$$ occurs twice while every other letter occurs once.

In the required list we include all six-letter arrangements of these letters except those in which the two $$R$$’s stand next to each other. Our task is to find how many valid words appear before the given word FARMER, and then add one for FARMER itself.

Step 1 : fixing the first letter. The first letter of FARMER is $$F$$. All words that start with letters alphabetically smaller than $$F$$ will certainly come before FARMER. The letters smaller than $$F$$ are $$A$$ and $$E$$ (two letters).

Suppose we fix such a smaller letter (say $$A$$) in the first place. We must then arrange the remaining five letters $$E,\,F,\,M,\,R,\,R$$ without letting the two $$R$$’s touch.

We first arrange the three non-$$R$$ letters $$E,\,F,\,M$$ in every possible order. Because they are all distinct, the number of such orders is $$3! = 6.$$

Whenever these three letters are written, they create $$4$$ “gaps’’ (one before the first letter, two between consecutive letters, and one after the last letter). To keep the two $$R$$’s apart, we simply choose any two of these four gaps and drop one $$R$$ into each chosen gap. Because the $$R$$’s are identical, the only decision is the choice of gaps, which can be done in $$\binom{4}{2} = 6$$ ways.

Hence the total number of valid words that start with a fixed smaller letter such as $$A$$ is $$3! \times \binom{4}{2} \;=\; 6 \times 6 \;=\; 36.$$

The same calculation holds if the first letter is $$E$$, so we obtain

$$36 + 36 = 72$$

valid words whose first letter precedes $$F$$. All those 72 words will appear before FARMER.

Step 2 : fixing the first two letters as $$F A$$. With the first letter now forced to be $$F$$, we turn to the second position. The second letter in FARMER is $$A$$. Among the unused letters $$A,\,E,\,M,\,R,\,R$$ none is alphabetically smaller than $$A$$, so no additional words are formed here. Our running count stays at $$72$$ words.

Step 3 : fixing the first three letters as $$F A$$ and choosing the third. The third letter in FARMER is $$R$$. The unused letters are $$E,\,M,\,R,\,R$$. Letters alphabetically smaller than $$R$$ are $$E$$ and $$M$$. We look at each choice in turn.

(i) Take $$E$$ as the third letter, giving the prefix $$F A E$$. The remaining letters are $$M,\,R,\,R$$. Among the $$3!/2! = 3$$ ordinary permutations of these three letters, only $$R M R$$ keeps the two $$R$$’s apart, so exactly one valid word is produced.

(ii) Take $$M$$ as the third letter, yielding the prefix $$F A M$$. The remaining letters are $$E,\,R,\,R$$, and again only $$R E R$$ is acceptable. Thus a second valid word is obtained.

Hence

$$1 + 1 = 2$$

additional words appear before any word that actually has $$F A R$$ as its first three letters. Adding these two words raises the count to

$$72 + 2 = 74.$$

Step 4 : fixing the prefix $$F A R$$ and choosing the fourth letter. The fourth letter in FARMER is $$M$$. After the prefix $$F A R$$ the unused letters are $$E,\,M,\,R$$. Among these, the letter alphabetically smaller than $$M$$ is only $$E$$.

If we select $$E$$ for the fourth place, the prefix becomes $$F A R E$$ and the leftover letters are $$M$$ and $$R$$. Because there is now only one $$R$$ remaining, there is no danger of placing two $$R$$’s together. The two distinct letters $$M$$ and $$R$$ can be arranged in

$$2! = 2$$

ways. Thus two more valid words precede FARMER, bringing the running total to

$$74 + 2 = 76.$$

Step 5 : fixing the prefix $$F A R M$$. With the fourth letter matching FARMER, the remaining letters are $$E$$ and $$R$$. The fifth letter in FARMER is $$E$$. Since no unused letter is alphabetically smaller than $$E$$, no new words are gained here.

The prefix now exactly matches $$F A R M E$$, leaving one final $$R$$ to be placed, which is allowed because it sits after an $$E$$ and is therefore not adjacent to another $$R$$.

Step 6 : counting FARMER itself. All the words counted so far (76 of them) precede FARMER. Consequently, the word FARMER occupies position

$$76 + 1 = 77.$$

So, the answer is $$77$$.