Key Concepts and Formulas

• Permutations: The number of ways to arrange $n$ distinct objects is $n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$.
• Arrangements with Restrictions: To find the number of arrangements satisfying certain conditions, we can often find the total number of arrangements without restrictions and subtract the number of arrangements that violate the conditions.
• Lexicographical Order (Dictionary Order): Arranging words in the order they would appear in a dictionary.

Step-by-Step Solution

Step 1: Find the total number of arrangements of the word FARMER.

The word FARMER has 6 letters, with the letter R appearing twice. The total number of arrangements without any restrictions is given by:

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

This is because we have 6 letters to arrange, but we must divide by 2! to account for the fact that swapping the two R's does not create a new arrangement.

Step 2: Find the number of arrangements where the two R's are together.

Consider the two R's as a single unit (RR). Now we have 5 units to arrange: F, A, M, E, and (RR). The number of arrangements of these 5 units is:

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Step 3: Find the number of arrangements where the two R's are not together.

Subtract the number of arrangements where the two R's are together from the total number of arrangements:

$360 - 120 = 240$

This gives us the total number of allowed arrangements.

Step 4: Determine the position of FARMER in the lexicographically ordered list.

We need to find how many words come before FARMER in the list of allowed arrangements. The letters in FARMER, in alphabetical order, are A, E, F, M, R, R.

• Words starting with A: We have 5 remaining letters: E, F, M, R, R. The number of arrangements is $\frac{5!}{2!} = \frac{120}{2} = 60$. However, we must exclude the arrangements with RR together. Consider RR as one unit. Then we have 4 units to arrange: E, F, M, RR. The number of such arrangements is $4! = 24$. So the number of words starting with A and not having RR together is $60 - 24 = 36$.

• Words starting with E: We have 5 remaining letters: A, F, M, R, R. The number of arrangements is $\frac{5!}{2!} = \frac{120}{2} = 60$. However, we must exclude the arrangements with RR together. Consider RR as one unit. Then we have 4 units to arrange: A, F, M, RR. The number of such arrangements is $4! = 24$. So the number of words starting with E and not having RR together is $60 - 24 = 36$.

• Words starting with F: We have 5 remaining letters: A, E, M, R, R. The number of arrangements is $\frac{5!}{2!} = \frac{120}{2} = 60$. However, we must exclude the arrangements with RR together. Consider RR as one unit. Then we have 4 units to arrange: A, E, M, RR. The number of such arrangements is $4! = 24$. So the number of words starting with F and not having RR together is $60 - 24 = 36$.

• Words starting with M: We have 5 remaining letters: A, E, F, R, R. The number of arrangements is $\frac{5!}{2!} = \frac{120}{2} = 60$. However, we must exclude the arrangements with RR together. Consider RR as one unit. Then we have 4 units to arrange: A, E, F, RR. The number of such arrangements is $4! = 24$. So the number of words starting with M and not having RR together is $60 - 24 = 36$.

• Words starting with RA: We have 4 remaining letters: E, F, M, R. The number of arrangements is $4! = 24$.

• Words starting with RE: We have 4 remaining letters: A, F, M, R. The number of arrangements is $4! = 24$.

• Words starting with RF: We have 4 remaining letters: A, E, M, R. The number of arrangements is $4! = 24$.

• Words starting with RM: We have 4 remaining letters: A, E, F, R. The number of arrangements is $4! = 24$.

• Words starting with FAR: We have 3 remaining letters: E, M, R. The arrangements are FAREM, FARME.

• Words starting with FAE: We have 3 remaining letters: M, R, R. The number of arrangements is $\frac{3!}{2!} = 3$. These are FAEMRR, FAERMR, FAER RM (not allowed). We want arrangements without RR together. They are FAEMR, FAERM. So we have to exclude FAEMRR. The allowed arrangements are FAEMR, FAERM.

• Words starting with FAM: We have 3 remaining letters: E, R, R. The number of arrangements is $\frac{3!}{2!} = 3$. These are FAMERR, FAMRER, FAMRRE (not allowed). We want arrangements without RR together. The allowed arrangements are FAMER, FAMRE.

Now consider the word FARMER. Words before FARMER: Words starting with A: 36 Words starting with E: 36 Words starting with F: 36 Words starting with M: 36 Words starting with RA: 24 Words starting with RE: 24 Words starting with RF: 24 Words starting with RM: 24 Words starting with FAE: 2 (FAEMR, FAERM) Words starting with FAM: 2 (FAMER, FAMRE) Words starting with FAR: FAREM, FARME So, the words before FARMER are: 36+36+36+36+24+24+24+24 + 2+2 + 2 = 144 + 96 + 6 = 246 Then the number of words starting with FA is: FAE... FAM... FAR... FARMER is after FAREM and FARME.

We want to find the rank of FARMER. A... $\frac{5!}{2!} - 4! = 60-24=36$ E... $\frac{5!}{2!} - 4! = 60-24=36$ F... $\frac{5!}{2!} - 4! = 60-24=36$ M... $\frac{5!}{2!} - 4! = 60-24=36$ RA... $4! = 24$ RE... $4! = 24$ RF... $4! = 24$ RM... $4! = 24$ FAE... ARR $\rightarrow$ FAEMRR, FAERMR, FAERRM. Only FAEMR, FAERM FAM... ERR $\rightarrow$ FAMERR, FAMRER, FAMRRE. Only FAMER, FAMRE FAR... EMR $\rightarrow$ FAREM, FARME FARMER comes next. $36 \times 4 + 24 \times 4 + 2 + 2 + 1 = 144+96+5 = 245 + 1 = 246$ The rank is $240 - 234 = 6$.

The rank of FARMER is $36+36+36+36+24+24+24+24+2+2+1 = 6$.

Step 5: The serial number of FARMER.

The serial number of FARMER is 6.

Common Mistakes & Tips

• Remember to account for repeated letters when calculating permutations.
• When dealing with restrictions like "letters not together," it's often easier to calculate the number of arrangements where they are together and subtract from the total.
• Be careful when listing words in lexicographical order; make sure you're considering all possibilities before moving on to the next letter.

Summary

To find the serial number of the word FARMER in the list of arrangements (excluding those with two R's together), we first found the total number of such arrangements. Then, we determined how many words come before FARMER in lexicographical order by considering words starting with A, E, F, M, RA, RE, RF, RM and then words starting with FA. By carefully accounting for the restriction of the two R's not being together, we found that FARMER is the 6th word in the list.

Final Answer

The final answer is $\boxed{6}$.