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Step-by-Step Solution
Step 1: Identify the problem
We want to list all distinct permutations of the letters of the word “FARMER” (i.e., the multiset {F, A, R, M, E, R}) under the condition that the two Rs do not appear together. Then, in alphabetical order, we must find the position (serial number) of “FARMER” in that list.
Step 2: List all possible letters in alphabetical order
The letters we have are:
A, E, F, M, R, R
We consider all valid permutations (with no two Rs together) and sort them alphabetically. Then we find where “FARMER” appears in this sorted list.
Step 3: Count how many valid permutations begin with a letter less than “F”
Letters less than “F” are A and E.
When the first letter is fixed as A, the remaining letters are {F, R, M, E, R}.
Total permutations of these 5 letters (with R repeated) is
\frac{5!}{2!} = 60 .
Permutations in which the two Rs are together (treat them as a single entity R*) would be
5 - 1 = 4 distinct symbols ⇒ 4! = 24 such permutations.
Hence, valid permutations (no two Rs together) starting with A = 60 - 24 = 36 .
Similarly, when the first letter is fixed as E, the remaining letters are {F, A, M, R, R}. By same reasoning:
Total permutations = \frac{5!}{2!} = 60 .
Those with Rs together = 4! = 24 .
Valid = 60 - 24 = 36 .
Thus, total valid permutations starting with a letter less than F = 36 + 36 = 72 .
Step 4: Fix the first letter as “F”
Now we look at the second letter. Our target word is “F A R M E R.” We count how many valid permutations start with “F” and have a second letter less than “A.” Since no letter is less than A, that count is zero.
Step 5: Fix the first two letters as “F, A”
The remaining letters are {R, M, E, R}. We look at the third letter of the word “FARMER,” which is “R,” and count how many valid permutations have third letter less than R.
Letters less than “R” (from {R, M, E, R}) are E and M.
Case 1: Third letter = E. The leftover set is {R, M, R}.
Number of distinct permutations = \frac{3!}{2!} = 3 .
Check for adjacency of Rs: only “RMR” is valid; “RRM” and “MRR” have two Rs together. So 1 valid permutation.
Case 2: Third letter = M. The leftover set is {R, E, R}.
Number of distinct permutations = \frac{3!}{2!} = 3 .
Only “RER” avoids consecutive Rs. So 1 valid permutation.
Therefore, when the first two letters are “F, A,” and the third letter is less than “R,” total valid permutations are 1 + 1 = 2.
Step 6: Fix the first three letters as “F, A, R”
The leftover letters are {R, M, E}, but effectively only one R is left. We now look at the fourth letter “M” from “FARMER” and count how many valid permutations have a fourth letter less than M.
Less than M in {R, M, E}? Alphabetically: E < M < R.
If the fourth letter = E, leftover = {M, R}.
Permutations of {M, R} = 2 (namely MR and RM). Neither has consecutive Rs (we have only 1 R left). So 2 valid permutations.
No other letter is less than M except E. So we have 2 such permutations.
Step 7: Assemble all permutations before “F, A, R, M, E, R”
Total permutations that come before “FARMER”:
72 (those starting with A or E),
+ 2 (third letter less than R after “F, A”),
+ 2 (fourth letter less than M after “F, A, R”).
Total = 72 + 2 + 2 = 76
Step 8: Conclude the position of “FARMER”
The word “FARMER” itself is the next one, so its rank = 76 + 1 = 77.
Final Answer
The serial number of the word “FARMER” in the list is 77.
