In the word RAINBOW, how many distinct permutations of all the letters are possible in which R and W occupy the two end positions (in either order)?

Introduction / Context:
This arrangement problem involves placing specific letters at fixed types of positions, namely the ends of a word. The word RAINBOW has seven distinct letters, and we must count how many permutations have R and W at the two extreme positions, in any order. Once the end letters are fixed, the remaining letters can be permuted freely in the middle positions.

Given Data / Assumptions:

• The word is RAINBOW, with letters R, A, I, N, B, O, W.
• All letters are distinct.
• The two end positions are the first and last positions of the seven letter arrangement.
• R and W must occupy the end positions in any order.
• The remaining five letters (A, I, N, B, O) fill the middle positions.

Concept / Approach:
First, determine how many ways R and W can be placed at the two ends. Then, determine how many ways to arrange the remaining five distinct letters in the five middle positions. Multiply these counts to get the total number of valid arrangements. This is a straightforward permutation problem with an additional constraint on specific positions.

Step-by-Step Solution:
Step 1: Place R and W at the two ends. The two end positions are left and right ends of the word. R can be at the left and W at the right, or W at the left and R at the right. Number of ways to assign R and W to the ends = 2. Step 2: Arrange the remaining letters in the middle positions. Remaining letters: A, I, N, B, O (5 distinct letters). Number of ways to arrange 5 distinct letters = 5! = 120. Step 3: Multiply the contributions. Total valid permutations = 2 * 5! = 2 * 120 = 240.

Verification / Alternative check:
Instead of thinking in two steps, we can imagine first choosing which letter goes to the left end (2 choices: R or W), which automatically fixes the right end as the other letter. Then we arrange the remaining 5 letters in the middle. This again leads to 2 * 120 = 240 permutations, confirming the result.

Why Other Options Are Wrong:
Option 120 corresponds to fixing R and W at specific ends but forgetting that they can be swapped. Option 180 or 210 do not match any simple product of 2 and a factorial. Option 360 would be 3 * 120, suggesting that three end configurations were counted, which is not possible here because only R and W can occupy the ends under the given condition.

Common Pitfalls:
A frequent mistake is to treat the two ends as indistinguishable, counting only 5! = 120 arrangements and ignoring swapping R and W. Another error is to forget that the remaining letters are all distinct and therefore should be arranged using 5!. Explicitly separating the assignment of end letters from the permutations of middle letters avoids such mistakes.

Final Answer:
There are 240 different permutations of RAINBOW where R and W occupy the two end positions.