To solve the problem of finding the number of permutations of the letters of the word "CONTRADICTORY" such that neither the pattern "CON" nor "RAD" nor "ORY" appears, we can use the principle of inclusion-exclusion. Here’s a step-by-step breakdown of the solution:
### Step 1: Calculate Total Arrangements
First, we need to find the total number of arrangements of the letters in the word "CONTRADICTORY".
The word "CONTRADICTORY" has 13 letters where:
- C appears 2 times,
- O appears 2 times,
- T appears 2 times,
- R appears 2 times,
- A, D, I, and Y each appear 1 time.
The formula for the total arrangements of letters is given by:
\[
\text{Total arrangements} = \frac{n!}{p_1! \cdot p_2! \cdot \ldots \cdot p_k!}
\]
Where \(n\) is the total number of letters, and \(p_i\) are the frequencies of the repeated letters.
Thus, we have:
\[
\text{Total arrangements} = \frac{13!}{2! \cdot 2! \cdot 2! \cdot 2!}
\]
### Step 2: Define Sets for Patterns
Let:
- \(A\) be the set of arrangements where "CON" appears.
- \(B\) be the set of arrangements where "RAD" appears.
- \(C\) be the set of arrangements where "ORY" appears.
We need to find the number of arrangements that do not belong to any of these sets, which is represented as \(A^c \cap B^c \cap C^c\).
### Step 3: Apply Inclusion-Exclusion Principle
According to the principle of inclusion-exclusion, we have:
\[
|A^c \cap B^c \cap C^c| = |U| - (|A| + |B| + |C| - |A \cap B| - |B \cap C| - |C \cap A| + |A \cap B \cap C|)
\]
Where \(|U|\) is the total arrangements calculated in Step 1.
### Step 4: Calculate Each Set Size
1. **Calculate \(|A|\)** (where "CON" appears together):
Treat "CON" as a single unit. The remaining letters are T, R, A, D, I, C, T, O, R, Y. This gives us 11 units (1 unit for "CON" + 10 other letters).
\[
|A| = \frac{11!}{2! \cdot 2! \cdot 2!}
\]
2. **Calculate \(|B|\)** (where "RAD" appears together):
Treat "RAD" as a single unit. The remaining letters are C, O, N, T, R, I, C, T, O, Y. This also gives us 11 units.
\[
|B| = \frac{11!}{2! \cdot 2! \cdot 2!}
\]
3. **Calculate \(|C|\)** (where "ORY" appears together):
Treat "ORY" as a single unit. The remaining letters are C, O, N, T, R, A, D, I, C, T. This again gives us 11 units.
\[
|C| = \frac{11!}{2! \cdot 2! \cdot 2!}
\]
4. **Calculate \(|A \cap B|\)** (where both "CON" and "RAD" appear):
Treat "CON" and "RAD" as single units. The remaining letters are T, C, I, T, O, R, Y. This gives us 9 units.
\[
|A \cap B| = \frac{9!}{2!}
\]
5. **Calculate \(|B \cap C|\)** (where both "RAD" and "ORY" appear):
Treat "RAD" and "ORY" as single units. The remaining letters are C, O, N, T, I, C. This gives us 9 units.
\[
|B \cap C| = \frac{9!}{2! \cdot 2!}
\]
6. **Calculate \(|C \cap A|\)** (where both "CON" and "ORY" appear):
Treat "CON" and "ORY" as single units. The remaining letters are T, R, A, D, I, C. This gives us 9 units.
\[
|C \cap A| = \frac{9!}{2! \cdot 2!}
\]
7. **Calculate \(|A \cap B \cap C|\)** (where "CON", "RAD", and "ORY" all appear):
Treat "CON", "RAD", and "ORY" as single units. The remaining letters are T, C, I. This gives us 7 units.
\[
|A \cap B \cap C| = 7!
\]
### Step 5: Substitute Values into Inclusion-Exclusion Formula
Now substitute all calculated values into the inclusion-exclusion formula:
\[
|A^c \cap B^c \cap C^c| = |U| - (|A| + |B| + |C| - |A \cap B| - |B \cap C| - |C \cap A| + |A \cap B \cap C|)
\]
### Step 6: Final Calculation
Calculate the final value using the above formula to get the number of arrangements that do not contain the patterns "CON", "RAD", or "ORY".
