All the arrangements, with or without meaning, of the word FARMER are written excluding any word tha

Q: Wait, why do we subtract the cases where the two $R$'s are together? Can't we just count the valid ones directly?

You absolutely could, but it would be much harder! Counting valid arrangements directly often requires complex case-work (e.g., placing $R$'s in the gaps between other letters). The complement method—Total minus Forbidden—is a powerful shortcut in combinatorics. It simplifies the problem significantly by turning a 'constraint' into a 'subtraction'.

Q: Why do we use $\frac{n!}{p!q!}$ for the total arrangements?

Great question! That formula is for permutations of a multiset. Since the word $FARMER$ has six letters, but the letter $R$ repeats twice, the total number of distinct arrangements is $\frac{6!}{2!}$. If all letters were unique, it would just be $6!$. Dividing by $2!$ removes the overcounting caused by swapping the two identical $R$'s.

Q: In the final steps, why does the constraint 'vanish' when we reach the prefix $FAR$?

Think about the letters remaining. Once we fix $F, A, R$ as the first three letters, we have used one $R$. The remaining letters are $E, M, R$. Since there is only one $R$ left in the pool, it is physically impossible to have two $R$'s together! The constraint is satisfied automatically for all remaining words.

Question

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 .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to master a classic problem in Permutations and Combinations. We are looking for the dictionary rank of the word $FARMER$, but with a twist: we must exclude any arrangement where the two $R$'s appear together.

Before we dive into the counting, we must organize our tools. The word is $FARMER$. First, let's list the letters and sort them alphabetically: $A, E, F, M, R, R$. This alphabetical order is the backbone of our dictionary.

Now, let's address the constraint: *No two $R$'s together*. As I mentioned in the hints, the most elegant way to handle this is the **Complement Method**. We calculate the total number of arrangements and subtract the 'forbidden' ones—the ones where the $R$'s are stuck together.

### The Master Equation

Total arrangements of the letters in $FARMER$ (6 letters, with $R$ repeating twice) is given by:

$$	ext{Total} = \frac{6!}{2!} = \frac{720}{2} = 360$$

Now, for the forbidden cases, we use the **Tie Method**. We treat the two $R$'s as a single unit, $(RR)$. Now, instead of 6 letters, we have 5 units: $A, E, F, M, (RR)$. The number of ways to arrange these 5 units is $5! = 120$.

So, the total number of valid words in our dictionary is:

$$	ext{Valid Words} = 360 - 120 = 240$$

### Building the Dictionary

Now, we systematically count the words that appear before $FARMER$.

**Words starting with $A$:**
If we fix $A$ at the start, we have 5 letters left: $E, F, M, R, R$. The total arrangements are $\frac{5!}{2!} = 60$. The forbidden cases (where $R$'s are together) are $4! = 24$. Thus, valid words starting with $A$ are $60 - 24 = 36$.

**Words starting with $E$:**
By symmetry, the logic is identical to the $A$ case. There are exactly $36$ valid words starting with $E$.

### The Final Stretch

Now we reach the words starting with $F$. Since our target word $FARMER$ starts with $F$, we must proceed letter by letter.

1. **Prefix $FAE$:** We have $M, R, R$ left. Total arrangements are $\frac{3!}{2!} = 3$. Forbidden cases (where $R$'s are together) are $2! = 2$. Valid words = $3 - 2 = 1$.

2. **Prefix $FAM$:** Similarly, we have $E, R, R$ left. The math is the same: $3 - 2 = 1$ valid word.

3. **Prefix $FAR$:** Here, the constraint vanishes! We have used one $R$, so only one $R$ remains. The remaining letters are $E, M, R$. The number of arrangements for the prefix $FARE$ is $2! = 2$.

4. **The Target:** After $FARE$, we move to $FARM$. The remaining letters are $E$ and $R$. The first word alphabetically is $FARMER$. This is our target!

### Final Calculation

Adding these up, we get:

$$	ext{Rank} = 36 (	ext{starts with } A) + 36 (	ext{starts with } E) + 1 (FAE) + 1 (FAM) + 2 (FARE) + 1 (FARMER) = 77$$

And there you have it! The **rank is 77**. It is all about breaking the problem down into manageable, logical steps. Keep practicing, and you will find that even the most complex problems become simple stories.

Enter Numerical Value:

Visualized Solution

Alphabetical Ordering of Letters

Understanding the Constraint

Calculating Total Arrangements

Arrangements with $R R$ Together

Total Valid Words in Dictionary

Words Starting with $A$

Words Starting with $E$

Words Starting with $F$ - Subcase $F A E$

Subcase $F A M$

Subcase $F A R$ - The Breakthrough

Reaching $F A R M E R$

Final Rank Calculation

The Sigma Insight: Linear Permutations

Analyzing the Setup

The Master Equation

Building the Dictionary

The Final Stretch

Final Calculation

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