If `(n!)/((2!)xx(n-2)!): (n!)/((4!)xx(n-4)!)=2:1`, find the value of n.

Explanation

To solve the problem, we start with the given equation:

\[ \frac{n!}{(2!) \cdot (n-2)!} : \frac{n!}{(4!) \cdot (n-4)!} = 2 : 1 \]

### Step 1: Simplify the Ratios

We can rewrite the left-hand side of the equation:

\[ \frac{n!}{(2!) \cdot (n-2)!} = \frac{n(n-1)}{2} \]

and

\[ \frac{n!}{(4!) \cdot (n-4)!} = \frac{n(n-1)(n-2)(n-3)}{24} \]

### Step 2: Set Up the Equation

Now, substituting these simplifications back into the ratio gives us:

\[ \frac{\frac{n(n-1)}{2}}{\frac{n(n-1)(n-2)(n-3)}{24}} = \frac{2}{1} \]

### Step 3: Cross-Multiply

Cross-multiplying gives us:

\[ \frac{n(n-1)}{2} \cdot 24 = 2 \cdot n(n-1)(n-2)(n-3) \]

### Step 4: Simplify the Equation

This simplifies to:

\[ 12n(n-1) = 2n(n-1)(n-2)(n-3) \]

Dividing both sides by \(2n(n-1)\) (assuming \(n \neq 0\) and \(n \neq 1\)) gives:

\[ 6 = (n-2)(n-3) \]

### Step 5: Expand the Right Side

Expanding the right side:

\[ n^2 - 5n + 6 = 6 \]

### Step 6: Rearranging the Equation

Rearranging gives:

\[ n^2 - 5n = 0 \]

### Step 7: Factor the Quadratic

Factoring out \(n\):

\[ n(n - 5) = 0 \]

### Step 8: Solve for \(n\)

Setting each factor to zero gives us:

\[ n = 0 \quad \text{or} \quad n = 5 \]

Since \(n\) must be a positive integer in the context of factorials, we take:

\[ n = 5 \]

### Final Answer

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

\[ \boxed{5} \]