We want to count the number of permutations of the letters $a, b, c, d, e$ that contain the substrings $ab$ or $bc$.

Step 1: Total permutations

The total number of permutations of the letters $a, b, c, d, e$ is: $5! = 120$

Step 2: Inclusion-exclusion principle

To find the number of permutations that contain $ab$ or $bc$, we use the inclusion-exclusion principle: $|A \cup B| = |A| + |B| - |A \cap B|$ where:

• $A$: Permutations that contain $ab$ as a substring.
• $B$: Permutations that contain $bc$ as a substring.

Step 3: Count $|A|$ (Permutations containing $ab$)

Treat $ab$ as a single block. Now, we are arranging the blocks $\{ab, c, d, e\}$, which gives: $4! = 24$ So, $|A| = 24$.

Step 4: Count $|B|$ (Permutations containing $bc$)

Similarly, treat $bc$ as a single block. Now, we are arranging the blocks $\{a, bc, d, e\}$, which also gives: $4! = 24$ So, $|B| = 24$.

Step 5: Count $|A \cap B|$ (Permutations containing both $ab$ and $bc$)

If both $ab$ and $bc$ appear, the substring $abc$ must appear as a single block. Now, we are arranging the blocks $\{abc, d, e\}$, which gives: $3! = 6$ So, $|A \cap B| = 6$.

Step 6: Apply inclusion-exclusion

Substitute the values into the formula: $|A \cup B| = |A| + |B| - |A \cap B|$ $|A \cup B| = 24 + 24 - 6 = 42$

Final Answer:

The number of permutations that contain $ab$ or $bc$ is: $\boxed{42}$

Let me know if you'd like further clarification or details! Here are some related questions to explore:

1. How many permutations contain both $ab$ and $de$?
2. What is the probability that a random permutation contains $bc$?
3. How many permutations contain $ab$ but not $bc$?
4. Can this approach generalize to strings longer than 5 letters? How?
5. How would the solution change if repetitions of letters were allowed?

Tip: When solving permutation problems, breaking the problem into blocks and using the inclusion-exclusion principle can simplify calculations significantly!