In how many different ways can the letters of the word "GLACIOUS" be arranged if the letter C must always occupy the last position?

Introduction / Context:
Here we work with permutations of letters under a positional restriction. All letters are distinct, but one specified letter must always appear in a particular fixed position. These kinds of questions reinforce the idea that once a position is fixed, we simply permute the remaining letters freely in the remaining places.

Given Data / Assumptions:

• Word: GLACIOUS.
• Letters are G, L, A, C, I, O, U, S (8 distinct letters).
• The letter C must always be placed at the last position of the arrangement.
• All remaining positions are filled by the other letters with no repetition.
• Order matters, so permutations are used.

Concept / Approach:
If one position is fixed for a particular letter, that letter does not contribute to the permutations among the remaining positions. We simply remove it from the pool and then count permutations of the remaining letters. For n distinct objects placed in n positions, the number of permutations is n!. So once the last position is fixed with C, we just permute the remaining 7 distinct letters in the first 7 positions.

Step-by-Step Solution:
Total letters in "GLACIOUS" = 8 distinct letters. The condition says that C must be at the last position of the arrangement. Fix C in the last position. Now there are 7 positions left and 7 remaining letters: G, L, A, I, O, U, S. The number of permutations of 7 distinct letters is 7!. Compute 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040.

Verification / Alternative check:
If there were no restriction, the number of permutations of 8 letters would be 8! = 40320. Since the letter C must always be at the last place, we are effectively counting how many of those 40320 arrangements have C at that position. Because all 8 positions are equally likely for C when there is no restriction, exactly 1 out of every 8 arrangements has C at the last position. So the count with the restriction should be 40320 / 8 = 5040, which matches our direct calculation. This gives a nice consistency check.

Why Other Options Are Wrong:

• 3360: This does not match any logical factorial based on 7 distinct letters. It suggests a miscalculation or a confusion with combinations.
• 720: This equals 6!, which would correspond to permuting only 6 letters instead of 7.
• 1080: This is also not a standard factorial and does not come from any correct model of the problem.

Common Pitfalls:
One common mistake is to forget that after fixing C at the last position, there are still 7 different letters left, not 6. Another issue is confusing permutations with combinations and attempting to use C(7,7), which only counts selections, not arrangements. Students may also try to first count all permutations and then divide incorrectly. The simplest method is to fix the constrained letter and then apply the factorial rule to the remaining letters.

Final Answer:
The number of arrangements of the letters of "GLACIOUS" with C fixed at the last position is 5040.