Loading...
The letters C, I, R, C, L, and E can be used to form 6-letter strings such as CIRCLE or CCIRLE. Using these letters, how many different 6-letter strings can be formed in which the two occurrences of the letter C are separated by at least one other letter?
Let's start by understanding what we're working with. We have 6 letters: C, I, R, C, L, and E. Notice that we have two C's and four other distinct letters.
The key constraint is that the two C's must be "separated by at least one other letter." This means the two C's cannot be next to each other in our 6-letter string. For example:
Process Skill: TRANSLATE - Converting the constraint "separated by at least one letter" into the mathematical concept of "non-adjacent positioning"
Let's think about this step by step. If all 6 letters were different, we'd have \(6!\) ways to arrange them. But since we have two identical C's, we need to account for this.
Imagine we arrange all 6 letters: _ _ _ _ _ _
If the letters were all different, each position could be filled in \(6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\) ways.
However, since the two C's are identical, any arrangement where we swap the positions of the two C's gives us the same string. For example, if C₁ is in position 1 and C₂ is in position 3, swapping them gives the exact same string as when C₂ is in position 1 and C₁ is in position 3.
Therefore, we need to divide by 2! to account for the identical C's:
Total arrangements = \(\frac{6!}{2!} = \frac{720}{2} = 360\)
Now let's count how many arrangements have the two C's next to each other. We can think of this by treating the two C's as a single "super letter" or unit.
If we glue the two C's together as one unit "CC", then we're arranging 5 objects:
These 5 objects can be arranged in \(5! = 120\) ways.
But wait - within the CC unit, the two C's can be arranged as CC or CC. Since the C's are identical, there's actually only 1 way to arrange them within the unit (CC is the same as CC).
So the number of arrangements with adjacent C's = \(5! \times 1 = 120\)
Now we use the complement principle. The arrangements where C's are separated equals the total arrangements minus the arrangements where C's are adjacent.
Let me put this in simple terms:
This makes logical sense - we're removing the "bad" arrangements (where C's touch) from all possible arrangements to get the "good" arrangements (where C's are separated).
The number of different 6-letter strings where the two C's are separated by at least one other letter is 240.
Looking at our answer choices:
Our answer matches choice E. 240.
1. Misinterpreting the constraint "separated by at least one other letter"
Students often misread this as "separated by exactly one letter" or "separated by more than one letter." The phrase "at least one" means one OR more letters between the C's, so CC is not allowed, but CIRCLE, CIRLCE, and ICRCEL are all valid since they have 1, 2, or 3 letters between the C's respectively.
2. Failing to recognize this as a complementary counting problem
Many students try to directly count arrangements where C's are separated by attempting to place C's in non-adjacent positions first, then arranging other letters. This approach becomes very complex and error-prone. The elegant solution is to use complementary counting: Total arrangements - Adjacent arrangements = Separated arrangements.
3. Not identifying that we have repeated letters (two C's)
Students may treat all six letters as distinct and calculate \(6! = 720\) total arrangements, forgetting that swapping the two identical C's doesn't create a new arrangement. This leads to overcounting and incorrect final answers.
1. Incorrectly calculating total arrangements with repeated letters
When calculating \(\frac{6!}{2!}\), students sometimes forget to divide by 2! for the repeated C's, or they incorrectly divide by other factorials. Some may also confuse this with the permutation formula nPr and use 6P6 instead of recognizing this as arrangements with repetition.
2. Mishandling the "treat CC as one unit" concept
When counting adjacent arrangements, students may incorrectly think the CC unit can be arranged internally in 2! = 2 ways (treating C's as distinguishable), leading to \(5! \times 2 = 240\) instead of the correct \(5! \times 1 = 120\). Since the C's are identical, there's only one way to arrange them within the unit.
3. Arithmetic errors in factorial calculations
Students may make basic computational errors like calculating \(5!\) as 60 instead of 120, or \(6!\) as 600 instead of 720. These errors compound through the problem, leading to incorrect intermediate results and wrong final answers.
1. Confusing intermediate results with the final answer
Students might select 120 (the number of adjacent arrangements) or 360 (the total arrangements) instead of 240 (the separated arrangements). This happens when they lose track of what the question is actually asking for after performing multiple calculations.