If the letters of the word SACHIN are arranged in all possible ways and written in dictionary order, at which serial number does the word SACHIN appear in that list?

Introduction / Context:
This question tests the idea of lexicographic or dictionary order of permutations of letters. Instead of listing every arrangement of the word SACHIN, which would be very time consuming, we use a systematic counting method based on factorials. Understanding this method is very important for many permutation and combination questions used in aptitude exams.

Given Data / Assumptions:
The word is SACHIN. The letters are S, A, C, H, I and N. All six letters are distinct. We assume standard dictionary order based on the English alphabet. Alphabetical order of the letters: A, C, H, I, N, S.

Concept / Approach:
The rank of a word in dictionary order is found by counting how many valid permutations come before it. For each position, we count how many smaller letters (in alphabetical order) could appear there, multiply that count by the number of permutations of the remaining positions, and add these contributions. Finally, we add 1 for the word itself, because the count of permutations before it does not include the word SACHIN itself.

Step-by-Step Solution:
Step 1: List letters in alphabetical order: A, C, H, I, N, S. Step 2: First letter is S. Letters smaller than S among {A, C, H, I, N, S} are A, C, H, I, N (5 letters). Step 3: For each such smaller letter at the first position, the remaining 5 letters can be arranged in 5! ways. So count = 5 * 5! = 5 * 120 = 600. Step 4: After fixing S at the first position, we move to the second letter A. Now remaining letters are A, C, H, I, N in alphabetical order, and A is the smallest among them. Step 5: There are no letters smaller than A among the remaining letters to place at the second position, so contribution here is 0. Step 6: Now fix SA and consider the third letter C. Remaining letters are C, H, I, N in alphabetical order. C is the smallest, so again contribution here is 0. Step 7: Fix SAC and look at H. Remaining letters are H, I, N; H is the smallest, so contribution again is 0. Step 8: Fix SACH and examine I. Remaining letters are I, N; I is the smaller, so contribution is 0. Step 9: Fix SACHI and then N is the only letter left, so we are at the word SACHIN. Step 10: Total permutations before SACHIN = 600. Therefore rank of SACHIN = 600 + 1 = 601.

Verification / Alternative check:
A quick check is to notice that S is the largest letter alphabetically. Hence all permutations starting with any of the other five letters (A, C, H, I, N) will come before any word starting with S. Since there are 5! words starting with each of those choices, the count of words before any S-starting word is exactly 5 * 5! = 600. Within the block of S-starting words, SACHIN is the first because the remaining letters A, C, H, I, N are already in alphabetical order. So the next rank after the first 600 permutations must be 601, which confirms our calculation.

Why Other Options Are Wrong:
600 would be the count of words strictly before SACHIN, not the rank of SACHIN itself. 602 and 603 overshoot the correct rank by adding too much. 604 is even further away and has no direct combinatorial justification here. Only 601 matches the correct method and reasoning for lexicographic rank with six distinct letters.

Common Pitfalls:
A very common mistake is to forget to add 1 at the end, thereby reporting the number of words before SACHIN instead of its rank. Another typical error is to assume that contributions will come from later positions even when the current letter is already the smallest among the remaining letters. Some students also wrongly consider that dictionary order is based on the word SACHIN itself rather than the alphabetical order of individual letters, which leads to confusion. Carefully ordering the letters and counting smaller letters at each step avoids these issues.

Final Answer:
Thus, the word SACHIN appears at serial number 601 when all permutations are listed in dictionary order.