How many ways can CROCODILE letters be split into three groups of three with the Cs separated?

Key takeaway: This is a partition-counting problem with a constraint on the two repeated C’s. The key is to count the arrangements of the 9 letter positions into 3 groups of 3, while forcing the two C’s into different groups.

We are dividing the 9 letter positions in CROCODILE into three groups of 3, and the two C’s must be in different groups.

Step 1: Count all partitions into 3 groups of 3

For 9 distinct letter positions, the number of ways to split them into 3 unlabeled groups of 3 is

$\frac{9!}{(3!)^3\,3!}=280.$

Step 2: Count the partitions where the two C’s are in the same group

It is easier to use the complement.

If the two C’s are together, choose the other 1 letter to join them:

$\binom{7}{1}=7$

Now the remaining 6 letters must be split into 2 groups of 3:

$\frac{6!}{(3!)^2\,2!}=10$

So the number with the two C’s together is

$7\times 10=70.$

Step 3: Subtract

Hence the number of ways with the two C’s in different groups is

$280-70=210.$

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Pitfalls the pros know 👇 The biggest trap is forgetting that the three groups are usually unlabeled. If you count them as labeled first, you must divide by $3!$ at the end. Another common miss is treating the two C’s as if they were already forced apart without checking the complementary case.

What if the problem changes? If the groups were labeled, the count would be larger: the same condition would give $1260$ labeled arrangements. If the requirement changed to “the two C’s must be in the same group,” the answer would be the complement, $70$.

Tags: set partitions, combinations, complement counting