If P(n) is the statement : `C_(r) le n!` for `1 le r le n` then:
(i) find `P(n + 1)`
(ii) show that `P(3)` is true

Answer

To solve the problem, we will follow the steps outlined in the question.

### Step 1: Find \( P(n + 1) \)

Given the statement \( P(n) \) is \( C_r \leq n! \) for \( 1 \leq r \leq n \), we need to express \( P(n + 1) \).

1. **Definition of \( P(n + 1) \)**:
\[ P(n + 1): C_r \leq (n + 1)! \]
for \( 1 \leq r \leq n + 1 \).

2. **Using the formula for combinations**:
The combination \( C_r \) can be expressed as:
\[ C_r = \frac{(n + 1)!}{r!(n + 1 - r)!} \]

3. **Inequality**:
To show that \( C_r \leq (n + 1)! \), we can rearrange the expression:
\[ \frac{(n + 1)!}{r!(n + 1 - r)!} \leq (n + 1)! \]

4. **Dividing both sides by \( (n + 1)! \)**:
This simplifies to:
\[ \frac{1}{r!(n + 1 - r)!} \leq 1 \]
which is true because \( r!(n + 1 - r)! \geq 1 \) for \( r \) in the range \( 1 \leq r \leq n + 1 \).

Thus, we have shown that \( P(n + 1) \) is true.

### Step 2: Show that \( P(3) \) is true

Now, we need to verify \( P(3) \).

1. **Definition of \( P(3) \)**:
\[ P(3): C_r \leq 3! \]
for \( 1 \leq r \leq 3 \).

2. **Calculate \( 3! \)**:
\[ 3! = 6 \]

3. **Calculate \( C_r \) for \( r = 1, 2, 3 \)**:
- For \( r = 1 \):
\[ C_1 = \frac{3!}{1!(3-1)!} = \frac{6}{1 \cdot 2} = 3 \quad \text{(which is } \leq 6\text{)} \]
- For \( r = 2 \):
\[ C_2 = \frac{3!}{2!(3-2)!} = \frac{6}{2 \cdot 1} = 3 \quad \text{(which is } \leq 6\text{)} \]
- For \( r = 3 \):
\[ C_3 = \frac{3!}{3!(3-3)!} = \frac{6}{6 \cdot 1} = 1 \quad \text{(which is } \leq 6\text{)} \]

4. **Conclusion**:
Since \( C_1, C_2, \) and \( C_3 \) are all less than or equal to \( 6 \), we conclude that \( P(3) \) is true.

### Summary of Results:
- \( P(n + 1) \) is true.
- \( P(3) \) is true.