How many 6-letter sequences can be formed that have 1 letter a, 2 letters b, and 3 letters c?

1. Translate the problem requirements: We need to find how many different ways we can arrange 6 letters where we have exactly 1 'a', exactly 2 'b's, and exactly 3 'c's in a sequence
2. Determine total positions and letter constraints: Identify that we have 6 positions to fill and verify that our letters (\(1+2+3=6\)) exactly fill all positions
3. Calculate arrangements using position selection: Choose positions for each letter type systematically - first select 1 position for 'a', then 2 positions for 'b's from remaining positions, then remaining positions automatically go to 'c's
4. Apply the multiplication principle: Multiply the number of ways to select positions for each letter type to get the total number of distinct sequences

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for. We want to create 6-letter sequences using specific letters with exact quantities:

• Exactly 1 letter 'a'
• Exactly 2 letters 'b'
• Exactly 3 letters 'c'

Think of this like having 6 empty boxes in a row: _ _ _ _ _ _

We need to place our letters in these boxes to create different sequences. For example, one possible sequence might be: a b b c c c. Another might be: c a c b c b.

The key insight is that we're counting how many different ways we can arrange these specific letters.

Process Skill: TRANSLATE - Converting the word problem into a clear arrangement task

2. Determine total positions and letter constraints

Let's verify our setup makes sense:

• Total positions available: 6
• Total letters we have: \(1 + 2 + 3 = 6\) ✓

Perfect! Our letters exactly fill all the positions, so every sequence will use all our letters.

Now, here's the crucial understanding: even though we have 2 b's, they are identical to each other. Same with the 3 c's - they're identical to each other. This means some arrangements that might seem different are actually the same sequence.

3. Calculate arrangements using position selection

Let's think about this step-by-step by deciding where each type of letter goes:

Step 1: Choose where the single 'a' goes
The letter 'a' can go in any of the 6 positions.
Number of ways = 6

Step 2: Choose where the 2 'b's go
After placing 'a', we have 5 remaining positions.
We need to choose 2 of these 5 positions for our 'b' letters.
Since the 2 'b's are identical, we just need to select which 2 positions they occupy.

To count this: imagine we have 5 positions left, and we're picking 2 of them.

• First 'b' position: 5 choices
• Second 'b' position: 4 remaining choices
• But since the b's are identical, we've double-counted! (picking position 2 then 4 is the same as picking position 4 then 2)
• So we divide by 2: \(\frac{5 \times 4}{2} = 10\)

Step 3: Place the 3 'c's
After placing 'a' and the 2 'b's, we have exactly 3 positions left.
The 3 'c's must go in these remaining positions.
Number of ways = 1 (no choice left)

4. Apply the multiplication principle

Now we multiply the number of ways for each step:

Total arrangements = (ways to place 'a') × (ways to place 'b's) × (ways to place 'c's)
Total arrangements = \(6 \times 10 \times 1 = 60\)

Let's double-check our calculation:

• 6 ways to choose position for 'a'
• 10 ways to choose 2 positions from remaining 5 for the 'b's
• 1 way to place 'c's in the remaining 3 positions
• \(6 \times 10 \times 1 = 60\)

Final Answer

The number of different 6-letter sequences we can form is 60.

This matches answer choice E.

To verify this makes sense: we started with what could have been \(6! = 720\) arrangements if all letters were different, but we divided out the repetitions caused by identical letters, giving us the much smaller number of 60 distinct sequences.

Common Faltering Points

Errors while devising the approach

• Treating identical letters as distinct: Students often fail to recognize that the 2 b's are identical to each other and the 3 c's are identical to each other. They might try to calculate \(6! = 720\) total arrangements, thinking each letter position creates a unique sequence, without accounting for the fact that swapping identical letters doesn't create a new sequence.
• Misunderstanding the arrangement vs. combination nature: Students may confuse this with a simple combination problem and try to use basic combination formulas without recognizing that this is specifically about arranging letters in sequence positions where order matters, but identical letters create constraints.
• Incorrectly applying permutation formulas: Some students might try to use basic permutation formulas like \(\mathrm{P}(6,3)\) or similar approaches without recognizing that this is a permutation with repetition problem that requires the multinomial coefficient approach or step-by-step position selection.

Errors while executing the approach

• Forgetting to account for identical letters when choosing positions: When calculating the ways to place 2 b's in 5 remaining positions, students often calculate \(5 \times 4 = 20\) but forget to divide by \(2! = 2\) to account for the fact that the two b's are identical, leading to an incorrect intermediate result of 20 instead of 10.
• Arithmetic errors in the multiplication principle: Students may correctly identify the individual steps (6 ways for 'a', 10 ways for 'b's, 1 way for 'c's) but make calculation errors when multiplying \(6 \times 10 \times 1\), potentially getting results like 600 or 16 instead of the correct 60.
• Double-counting or under-counting arrangements: Students might incorrectly calculate the number of ways to choose positions for identical letters, such as using \(\mathrm{C}(5,2) = 10\) correctly for the b's but then incorrectly trying to apply additional permutations or failing to recognize that the c's automatically fill the remaining positions.

Errors while selecting the answer

No likely faltering points