The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT....

1. Translate the problem requirements: We need to count 5-letter arrangements of D, G, I, I, T where the two I's are NOT adjacent (separated by at least one other letter)
2. Calculate total possible arrangements: Find all ways to arrange these 5 letters where two letters are identical
3. Count arrangements with adjacent I's: Determine how many arrangements have the two I's next to each other
4. Apply complementary counting: Subtract adjacent arrangements from total arrangements to get the final answer

Execution of Strategic Approach

1. Translate the problem requirements

Let's understand what we're looking for. We have 5 letters: D, G, I, I, and T. Notice that we have two I's and the other letters appear once each.

We want to arrange these 5 letters into strings where the two I's are NOT next to each other - they must be "separated by at least one other letter." This means there must be at least one letter between the two I's.

For example:

• DIGIT works (I's separated by G)
• DGIIT does NOT work (I's are adjacent)
• TIGDI works (I's separated by G and D)

Process Skill: TRANSLATE - Converting the constraint "separated by at least one other letter" into "not adjacent"

2. Calculate total possible arrangements

First, let's find how many ways we can arrange all 5 letters without any restrictions.

If all letters were different, we'd have \(5! = 120\) arrangements. However, since we have two identical I's, we need to account for this repetition.

When we have identical objects, we divide by the number of ways to arrange the identical items among themselves. The two I's can be arranged in \(2! = 2\) ways, but since they're identical, these arrangements look the same.

Total arrangements = \(\frac{5!}{2!} = \frac{120}{2} = 60\)

3. Count arrangements with adjacent I's

Now let's count how many arrangements have the two I's next to each other (adjacent).

We can think of the two adjacent I's as a single "block" or unit. So instead of arranging D, G, I, I, T, we're arranging D, G, [II], T - that's 4 units.

These 4 units can be arranged in \(4! = 24\) ways.

Let's verify with a few examples:

• [II]DGT → IIDGT
• D[II]GT → DIIGT
• DG[II]T → DGIIT
• DGT[II] → DGTII

So there are 24 arrangements where the I's are adjacent.

4. Apply complementary counting

To find arrangements where I's are NOT adjacent, we subtract the adjacent arrangements from the total arrangements.

Arrangements where I's are separated = Total arrangements - Adjacent arrangements

Arrangements where I's are separated = \(60 - 24 = 36\)

Process Skill: APPLY CONSTRAINTS - Using complementary counting to handle the "at least one letter between" constraint efficiently

4. Final Answer

The number of 5-letter strings where the two I's are separated by at least one other letter is 36.

This matches answer choice (D) 36.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misinterpreting the constraint "separated by at least one other letter"

Students often confuse this phrase and think it means the I's must have exactly one letter between them, rather than understanding it means "not adjacent" (one OR more letters between them). This leads to a completely different counting approach where they try to place exactly one letter between the I's.

Faltering Point 2: Forgetting to account for identical letters in permutation calculations

When calculating total arrangements, students frequently use \(5! = 120\) directly, forgetting that having two identical I's means we must divide by 2! to avoid overcounting. This gives an incorrect total of 120 instead of 60, making all subsequent calculations wrong.

Faltering Point 3: Attempting direct counting instead of using complementary counting

Students often try to directly count arrangements where I's are separated by considering all possible positions, which becomes complex and error-prone. They miss that complementary counting (Total - Adjacent) is much simpler and more reliable for this type of constraint.

Errors while executing the approach

Faltering Point 1: Incorrect calculation when treating adjacent I's as a block

When using the "block method" to count adjacent arrangements, students sometimes forget that the two I's within the block are identical, so they don't need to multiply by 2!. The correct count is simply \(4! = 24\) arrangements of the 4 units (D, G, [II], T).

Faltering Point 2: Arithmetic errors in the complementary counting subtraction

Students may correctly calculate total arrangements (60) and adjacent arrangements (24) but make simple arithmetic mistakes when subtracting: 60 - 24. Some might accidentally add instead of subtract, or make computational errors leading to incorrect final results.

Faltering Point 3: Errors while selecting the answer

No likely faltering points - the final calculation directly gives a single numerical result that matches one of the answer choices, with no additional interpretation or selection criteria needed.