Find the number of ways in which the 9 letters in the word CROCODILE can be divided into
three groups, each containing three letters, if the two Cs must be in different groups.

To solve the problem of finding the number of ways to divide the 9 letters in "CROCODILE" into three groups, each containing three letters, with the condition that the two Cs must be in different groups, we can approach it step by step.

### Step 1: Consider the Total Number of Groupings
First, let's find the total number of ways to divide the 9 letters into three groups of three without considering the condition on the Cs. The formula for this is given by:
\[
\text{Total groupings} = \frac{\binom{9}{3} \cdot \binom{6}{3}}{3!}
\]
where:
- \(\binom{9}{3}\) represents the number of ways to choose the first group of 3 letters from 9 letters.
- \(\binom{6}{3}\) represents the number of ways to choose the second group of 3 letters from the remaining 6 letters.
- \(3!\) (which equals 6) accounts for the fact that the order of the three groups doesn't matter.

Let's calculate this:
\[
\text{Total groupings} = \frac{\binom{9}{3} \cdot \binom{6}{3}}{6} = \frac{\frac{9!}{3! \cdot 6!} \cdot \frac{6!}{3! \cdot 3!}}{6} = \frac{\frac{9 \times 8 \times 7}{6} \cdot \frac{6 \times 5 \times 4}{6}}{6} = \frac{84 \cdot 20}{6} = 280
\]
So, there are 280 ways to divide the 9 letters into three groups of three.

### Step 2: Consider the Condition on the Cs
Now, we need to ensure that the two Cs are in different groups. 

1. **Choosing the groups for the Cs:** The first C can go into any of the 3 groups, and the second C can go into any of the remaining 2 groups. So, there are \(3 \times 2 = 6\) ways to place the Cs into different groups.

2. **Dividing the remaining 7 letters:** Once the Cs are placed, we are left with the 7 letters "R, O, O, D, I, L, E". These must be divided into the remaining three groups of three letters, with one letter already in each group (due to the placement of the Cs).

For the remaining 7 letters, one from each group has already been used, so we effectively have to distribute 7 letters across the 3 groups. This is the same as determining the number of groupings for 7 items, which again will follow a similar binomial approach.

Since the problem is symmetrical in groups and all groupings are equally likely:
\[
\text{Valid groupings} = \frac{6 \cdot \frac{\binom{7}{2} \cdot \binom{5}{2}}{2!}}{1} 
\]
But since we've already calculated the base case without Cs (280) and Cs can be placed in 6 ways, we simply multiply:
\[
\text{Total valid groupings} = \frac{6}{6} \times 280 = 280
\]

### Step 3: Adjust for Duplicate Letters
The word "CROCODILE" has duplicates:
- 2 Cs
- 2 Os

This reduces the total number of permutations by dividing by the factorial of the number of duplicates:
\[
\text{Final Answer} = \frac{280}{2! \cdot 2!} = \frac{280}{4} = 70
\]
Thus, the number of ways to divide the 9 letters into three groups, with the Cs in different groups, is **70**.

---

Do you want more details or have any questions? 

Here are some related questions:
1. How does the inclusion of identical letters affect the total number of groupings?
2. What would happen if the condition was that both Cs had to be in the same group?
3. How would the solution change if the word had no repeated letters?
4. What if we wanted to divide the letters into four groups instead of three?
5. How can this problem be solved using a combinatorial counting argument?

**Tip:** When dealing with combinatorial problems, always account for identical items separately, as they reduce the total number of distinct arrangements.

Explore the solution to dividing the letters in CROCODILE into three groups of three, ensuring the two Cs are in different groups. Learn about binomial coefficients and permutations used in combinatorial counting. Suitable for advanced high school students.

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