If `P + P! = P^2` then the value of P is:

Options

A. 5
B. 7
C. 3
D. 0

Answer

To solve the equation \( P + P! = P^2 \), we will analyze the equation step by step.

### Step 1: Understand the equation
We have the equation:
\[ P + P! = P^2 \]
This means that the sum of \( P \) and the factorial of \( P \) equals \( P \) squared.

### Step 2: Rearrange the equation
We can rearrange the equation to isolate \( P! \):
\[ P! = P^2 - P \]

### Step 3: Factor the right-hand side
We can factor the right-hand side:
\[ P! = P(P - 1) \]

### Step 4: Check integer values for \( P \)
Since \( P! \) is defined for non-negative integers, we will check integer values of \( P \) to see which satisfies the equation.

#### Check \( P = 3 \):
\[ 3! = 6 \quad \text{and} \quad 3(3 - 1) = 3 \times 2 = 6 \]
Thus, \( 3 + 6 = 9 \) and \( 3^2 = 9 \). This works.

#### Check \( P = 4 \):
\[ 4! = 24 \quad \text{and} \quad 4(4 - 1) = 4 \times 3 = 12 \]
Thus, \( 4 + 24 = 28 \) and \( 4^2 = 16 \). This does not work.

#### Check \( P = 5 \):
\[ 5! = 120 \quad \text{and} \quad 5(5 - 1) = 5 \times 4 = 20 \]
Thus, \( 5 + 120 = 125 \) and \( 5^2 = 25 \). This does not work.

#### Check \( P = 7 \):
\[ 7! = 5040 \quad \text{and} \quad 7(7 - 1) = 7 \times 6 = 42 \]
Thus, \( 7 + 5040 = 5047 \) and \( 7^2 = 49 \). This does not work.

### Conclusion
The only value of \( P \) that satisfies the equation \( P + P! = P^2 \) is:
\[ \boxed{3} \]