Key Concepts and Formulas

• Permutations: The number of ways to arrange $n$ distinct objects is $n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$.
• Principle of Inclusion-Exclusion (PIE) for Two Sets: $|A \cup B| = |A| + |B| - |A \cap B|$, where $|A \cup B|$ represents the number of elements in set A or set B or both, $|A|$ is the number of elements in set A, $|B|$ is the number of elements in set B, and $|A \cap B|$ is the number of elements in both set A and set B.
• Complement Rule: The number of elements not in a set A is given by $|U| - |A|$, where U is the universal set.

Step-by-Step Solution

Step 1: Calculate the Total Number of Permutations

We want to find the total number of permutations of the digits 1, 2, 3, 4, 5, 6, and 7 without any restrictions. Since there are 7 distinct digits, the total number of permutations is 7!.

$$\text{Total Permutations} = 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$$

This value represents the size of our universal set, $U$.

Step 2: Calculate |A| (Permutations containing "153")

Let $A$ be the set of permutations that contain the string "153". We use the block method, treating "153" as a single unit. This means we arrange the block "153" along with the digits 2, 4, 6, and 7. We now have 5 items to arrange: (153), 2, 4, 6, 7. The number of ways to arrange these 5 items is 5!.

$$|A| = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$

Step 3: Calculate |B| (Permutations containing "2467")

Let $B$ be the set of permutations that contain the string "2467". Again, we use the block method, treating "2467" as a single unit. We arrange the block "2467" along with the digits 1, 3, and 5. We now have 4 items to arrange: (2467), 1, 3, 5. The number of ways to arrange these 4 items is 4!.

$$|B| = 4! = 4 \times 3 \times 2 \times 1 = 24$$

Step 4: Calculate |A $\cap$ B| (Permutations containing both "153" and "2467")

$A \cap B$ represents the set of permutations that contain both the string "153" and the string "2467". Since the digits in "153" and "2467" are distinct, we can treat both as single units. We arrange the two blocks: (153) and (2467). The number of ways to arrange these 2 items is 2!.

$$|A \cap B| = 2! = 2 \times 1 = 2$$

Step 5: Apply the Principle of Inclusion-Exclusion for $|A \cup B|$

We use the PIE formula to find the number of permutations that contain at least one of the forbidden strings.

$$|A \cup B| = |A| + |B| - |A \cap B| = 120 + 24 - 2 = 142$$

Step 6: Find the Number of Permutations that Contain Neither String

We want to find the number of permutations that do not contain "153" and do not contain "2467". This is the total number of permutations minus the number of permutations that contain at least one of the forbidden strings.

$$\text{Number of Permutations with Neither String} = \text{Total Permutations} - |A \cup B| = 5040 - 142 = 4898$$

Common Mistakes & Tips

• Forgetting to subtract the intersection: A common mistake is to calculate $|A| + |B|$ but forget to subtract $|A \cap B|$ when using PIE.
• Incorrectly Calculating Factorials: Make sure to calculate the factorials correctly. It's easy to make a mistake in the multiplication.
• Overlapping Digits: If the strings shared digits, the calculation of $|A \cap B|$ would be more complex and involve considering how the blocks merge.

Summary

We used the Principle of Inclusion-Exclusion to find the number of permutations of the digits 1 through 7 that contain neither the string "153" nor the string "2467". We found the total number of permutations, then subtracted the number of permutations containing at least one of the forbidden strings. This gave us the final answer of 4898.

The final answer is \boxed{4898}.