The appropriate value of P for the relation `(P! + 1) = (P + 1)^2` is :

Options

A. 3
B. 4
C. 5
D. none of these

Answer

To solve the equation \( P! + 1 = (P + 1)^2 \), we will check the values of \( P \) given in the options.

### Step 1: Check \( P = 3 \)

1. Calculate \( 3! + 1 \):
\[ 3! = 3 \times 2 \times 1 = 6 \]
\[ 3! + 1 = 6 + 1 = 7 \]

2. Calculate \( (3 + 1)^2 \):
\[ (3 + 1)^2 = 4^2 = 16 \]

3. Compare:
\[ 7 \neq 16 \]
So, \( P = 3 \) is not a solution.

### Step 2: Check \( P = 4 \)

1. Calculate \( 4! + 1 \):
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]
\[ 4! + 1 = 24 + 1 = 25 \]

2. Calculate \( (4 + 1)^2 \):
\[ (4 + 1)^2 = 5^2 = 25 \]

3. Compare:
\[ 25 = 25 \]
So, \( P = 4 \) is a solution.

### Step 3: Check \( P = 5 \)

1. Calculate \( 5! + 1 \):
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
\[ 5! + 1 = 120 + 1 = 121 \]

2. Calculate \( (5 + 1)^2 \):
\[ (5 + 1)^2 = 6^2 = 36 \]

3. Compare:
\[ 121 \neq 36 \]
So, \( P = 5 \) is not a solution.

### Conclusion

The appropriate value of \( P \) for the relation \( P! + 1 = (P + 1)^2 \) is \( P = 4 \).

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