To solve this, let's break down the problem.

We are given that the permutations of the letters $ABCDEFG$ must contain the substrings "ABC" and "CDEF".

Step 1: Treat the substrings as single blocks

• The string "ABC" appears as a single block.
• The string "CDEF" also appears as a single block.

However, we see that "ABC" and "CDEF" share the letter "C", so when we combine them, we need to account for this overlap.

Thus, the blocks are:

1. $\text{"ABC"}$ (a block of three letters).
2. $\text{"DEF"}$ (a block of three letters, since "C" is already accounted for).

Step 2: Treat the remaining letters

After accounting for "ABC" and "DEF", the only letter left in the set $\{ A, B, C, D, E, F, G \}$ is $G$.

Step 3: Permute the blocks

Now we have three units to permute:

1. The block $\text{"ABC"}$.
2. The block $\text{"DEF"}$.
3. The letter $G$.

We are essentially permuting these 3 items.

Step 4: Compute the number of permutations

There are $3! = 6$ ways to permute the blocks "ABC", "DEF", and $G$.

Final Answer:

There are $6$ permutations of the letters $ABCDEFG$ that contain the substrings "ABC" and "CDEF".

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Would you like further clarification on any part of the solution?

Here are 5 related questions:

1. How many permutations contain the strings "AB" and "DE" as blocks?
2. What if there were no overlapping letters between the blocks "ABC" and "DEF"?
3. How many permutations contain only the string "ABC" as a block?
4. What if we allow other letters to be repeated within the permutation?
5. How many permutations of the letters "ABCDEFG" exist without any restrictions?

Tip: When solving problems with overlapping substrings, always account for shared letters carefully to avoid overcounting or undercounting blocks.