Find n if :
P(2n, 3) = 100 P(n,2)

Answer

To solve the equation \( P(2n, 3) = 100 P(n, 2) \), we will use the formula for permutations, which is given by:

\[ P(n, r) = \frac{n!}{(n-r)!} \]

### Step 1: Write the permutation expressions

Using the formula for permutations, we can express both sides of the equation:

\[ P(2n, 3) = \frac{(2n)!}{(2n-3)!} \]
\[ P(n, 2) = \frac{n!}{(n-2)!} \]

### Step 2: Substitute the expressions into the equation

Now substituting these expressions into the original equation, we have:

\[ \frac{(2n)!}{(2n-3)!} = 100 \cdot \frac{n!}{(n-2)!} \]

### Step 3: Simplify both sides

The left-hand side simplifies to:

\[ (2n)(2n-1)(2n-2) \]

So the equation becomes:

\[ (2n)(2n-1)(2n-2) = 100 \cdot (n)(n-1) \]

### Step 4: Expand both sides

Expanding both sides gives:

\[ (2n)(2n-1)(2n-2) = 8n^3 - 12n^2 + 4n \]
\[ 100(n)(n-1) = 100(n^2 - n) = 100n^2 - 100n \]

### Step 5: Set the equation to zero

Now we set the equation to zero:

\[ 8n^3 - 12n^2 + 4n - 100n^2 + 100n = 0 \]

Combining like terms yields:

\[ 8n^3 - 112n^2 + 104n = 0 \]

### Step 6: Factor out common terms

Factoring out \( 8n \):

\[ 8n(n^2 - 14n + 13) = 0 \]

### Step 7: Solve the quadratic equation

Now we solve the quadratic equation \( n^2 - 14n + 13 = 0 \) using the quadratic formula:

\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here \( a = 1, b = -14, c = 13 \):

\[ n = \frac{14 \pm \sqrt{(-14)^2 - 4 \cdot 1 \cdot 13}}{2 \cdot 1} \]
\[ = \frac{14 \pm \sqrt{196 - 52}}{2} \]
\[ = \frac{14 \pm \sqrt{144}}{2} \]
\[ = \frac{14 \pm 12}{2} \]

This gives us two potential solutions:

\[ n = \frac{26}{2} = 13 \quad \text{and} \quad n = \frac{2}{2} = 1 \]

### Step 8: Verify the solutions

Since \( n \) must be a positive integer, we check both values in the context of the problem. Both values are valid, but we were asked for \( n \) in the context of the original equation.

### Final Answer

Thus, the value of \( n \) is:

\[ \boxed{13} \]