Find n if :
` 2P( n,3) = P(n+1,3)`

Answer

To solve the equation \( 2P(n, 3) = P(n+1, 3) \), we will follow these steps:

### Step 1: Write the permutation formulas
The permutation formula is given by:
\[ P(n, r) = \frac{n!}{(n-r)!} \]
Thus, we can express \( P(n, 3) \) and \( P(n+1, 3) \) as follows:
\[ P(n, 3) = \frac{n!}{(n-3)!} \]
\[ P(n+1, 3) = \frac{(n+1)!}{(n+1-3)!} = \frac{(n+1)!}{(n-2)!} \]

### Step 2: Substitute the permutations into the equation
Now substitute these expressions into the original equation:
\[ 2 \cdot \frac{n!}{(n-3)!} = \frac{(n+1)!}{(n-2)!} \]

### Step 3: Simplify the equation
We can simplify \( (n+1)! \) as follows:
\[ (n+1)! = (n+1) \cdot n! \]
So the equation becomes:
\[ 2 \cdot \frac{n!}{(n-3)!} = \frac{(n+1) \cdot n!}{(n-2)!} \]

### Step 4: Cancel \( n! \) from both sides
Assuming \( n! \neq 0 \) (which is valid for \( n \geq 3 \)), we can cancel \( n! \):
\[ 2 \cdot \frac{1}{(n-3)!} = \frac{(n+1)}{(n-2)!} \]

### Step 5: Rewrite \( (n-2)! \) in terms of \( (n-3)! \)
We know that:
\[ (n-2)! = (n-2)(n-3)! \]
Substituting this into the equation gives:
\[ 2 \cdot \frac{1}{(n-3)!} = \frac{(n+1)}{(n-2)(n-3)!} \]

### Step 6: Cancel \( (n-3)! \) from both sides
Now we can cancel \( (n-3)! \) from both sides:
\[ 2 = \frac{(n+1)}{(n-2)} \]

### Step 7: Cross-multiply to solve for \( n \)
Cross-multiplying gives:
\[ 2(n-2) = n + 1 \]
Expanding this results in:
\[ 2n - 4 = n + 1 \]

### Step 8: Rearrange the equation
Rearranging gives:
\[ 2n - n = 4 + 1 \]
\[ n = 5 \]

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{5} \]