Number of distinct arrangements of the word SCOOTY when letters S and Y are fixed at the two ends

Introduction / Context:
This problem checks your understanding of permutations of letters in a word when certain positions are fixed. The word SCOOTY contains repeating letters, so you must apply the permutation formula for objects with repetition, combined with positional restrictions.

Given Data / Assumptions:

• Word: SCOOTY.
• Letters: S, C, O, O, T, Y.
• The letters S and Y must always occupy the two end positions in some order.
• The remaining letters C, O, O, T fill the middle four positions.
• All arrangements are case insensitive and every arrangement counts as a distinct word even if it has no meaning.

Concept / Approach:
We fix which letters go at the ends and then arrange the remaining letters in the middle positions. Because there is a repeated letter O, we use the formula for permutations of n objects with repetition: n factorial divided by the product of factorials of the counts of identical objects.

Step-by-Step Solution:
Step 1: The two ends must be S and Y. They can appear as S at the left and Y at the right or Y at the left and S at the right.Step 2: Number of ways to arrange S and Y at the two ends = 2! = 2.Step 3: The middle four positions must be filled with C, O, O, T.Step 4: If all four letters were distinct, we would have 4! arrangements.Step 5: But O appears twice, so the count must be divided by 2! to correct for repetition.Step 6: Number of arrangements of C, O, O, T = 4! / 2! = 24 / 2 = 12.Step 7: Total valid arrangements = ways to arrange the ends * ways to arrange the middle = 2 * 12 = 24.

Verification / Alternative check:
You can explicitly imagine fixing S on the left and Y on the right, then list or count the permutations of C, O, O, T to confirm there are 12. Repeating the same with Y on the left and S on the right would again give 12, so the total is 24. This confirms the formula based approach.

Why Other Options Are Wrong:

• 720 would be the number of arrangements of 6 distinct letters, which does not apply due to repetition and fixed ends.
• 360 is still too large and does not match 2 * 4! or any correct adjustment for repeated letters.
• 120 equals 5! and would correspond to an incorrect assumption about only 5 letters being arranged freely.

Common Pitfalls:
Students often ignore the repeated letter O and simply compute 2 * 4!, giving 48, which is double the correct answer. Sometimes, they may also forget that S and Y can swap ends, leading them to count only 12 arrangements. Always separate the problem into fixed positions and free positions and then apply the permutation rules carefully, especially when letters repeat.

Final Answer:
The number of valid arrangements is 24, so the correct answer is 24.