Find n if :
` ""^(n)P_2 = 30`

Answer

To solve the problem, we need to find the value of \( n \) such that \( {}^nP_2 = 30 \).

### Step 1: Understand the formula for permutations
The formula for permutations of \( n \) items taken \( r \) at a time is given by:
\[ {}^nP_r = \frac{n!}{(n-r)!} \]
In our case, \( r = 2 \), so we have:
\[ {}^nP_2 = \frac{n!}{(n-2)!} \]

### Step 2: Set up the equation
From the problem, we know:
\[ {}^nP_2 = 30 \]
Thus, we can write:
\[ \frac{n!}{(n-2)!} = 30 \]

### Step 3: Simplify the equation
We can simplify \( \frac{n!}{(n-2)!} \) as follows:
\[ \frac{n!}{(n-2)!} = n \times (n-1) \]
So, we rewrite our equation:
\[ n(n-1) = 30 \]

### Step 4: Rearrange the equation
Now, we can rearrange the equation:
\[ n^2 - n - 30 = 0 \]

### Step 5: Factor the quadratic equation
Next, we need to factor the quadratic equation:
\[ n^2 - n - 30 = (n - 6)(n + 5) = 0 \]

### Step 6: Solve for \( n \)
Setting each factor to zero gives us:
\[ n - 6 = 0 \quad \Rightarrow \quad n = 6 \]
\[ n + 5 = 0 \quad \Rightarrow \quad n = -5 \]
Since \( n \) must be a non-negative integer, we discard \( n = -5 \).

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{6} \]