Let P (n) be the statement `C_(r) le n!` for `1 le r le n` then `P(n + 1)` = ________

Answer

To solve the problem, we need to express the statement \( P(n+1) \) based on the given statement \( P(n) \). The statement \( P(n) \) is defined as:

\[ P(n): C_r \leq n! \]

for \( 1 \leq r \leq n \).

### Step-by-step Solution:

1. **Understanding the Binomial Coefficient**:
The binomial coefficient \( C_r \) is defined as:
\[ C_r = \frac{n!}{r!(n-r)!} \]
Therefore, the statement \( P(n) \) can be rewritten as:
\[ \frac{n!}{r!(n-r)!} \leq n! \]

2. **Rewriting \( P(n+1) \)**:
Now, we need to express \( P(n+1) \). The statement becomes:
\[ P(n+1): C_r \leq (n+1)! \]
for \( 1 \leq r \leq n+1 \).

3. **Expressing \( C_r \) for \( n+1 \)**:
The binomial coefficient for \( n+1 \) is:
\[ C_r = \frac{(n+1)!}{r!(n+1-r)!} \]

4. **Comparing \( C_r \) with \( (n+1)! \)**:
We need to show that:
\[ \frac{(n+1)!}{r!(n+1-r)!} \leq (n+1)! \]
This simplifies to:
\[ \frac{1}{r!(n+1-r)!} \leq 1 \]

5. **Understanding the Inequality**:
The inequality \( \frac{1}{r!(n+1-r)!} \leq 1 \) holds true for \( 1 \leq r \leq n+1 \) because \( r! \) and \( (n+1-r)! \) are both positive integers.

6. **Conclusion**:
Therefore, we conclude that:
\[ P(n+1): C_r \leq (n+1)! \]
for \( 1 \leq r \leq n+1 \).

### Final Answer:
Thus, \( P(n+1) \) can be expressed as:
\[ P(n+1): C_r \leq (n+1)! \]