If ` (n!)/( 2!( n-2) !) and (eta !)/( 4!(n-4)!) ` are in the ratio 2: 1 , find the value of n.

Answer

To solve the problem, we need to find the value of \( n \) given that the expressions \( \frac{n!}{2!(n-2)!} \) and \( \frac{n!}{4!(n-4)!} \) are in the ratio \( 2:1 \).

### Step-by-Step Solution:

1. **Write the given ratio**:
\[ \frac{\frac{n!}{2!(n-2)!}}{\frac{n!}{4!(n-4)!}} = \frac{2}{1} \]

2. **Simplify the ratio**:
The \( n! \) in the numerator and denominator cancels out:
\[ \frac{1}{2!(n-2)!} \cdot \frac{4!(n-4)!}{1} = 2 \]

3. **Substituting the factorial values**:
We know \( 2! = 2 \) and \( 4! = 24 \):
\[ \frac{24(n-4)!}{2(n-2)!} = 2 \]

4. **Multiply both sides by \( 2(n-2)! \)**:
\[ 24(n-4)! = 4(n-2)! \]

5. **Express \( (n-2)! \) in terms of \( (n-4)! \)**:
We can write \( (n-2)! \) as \( (n-2)(n-3)(n-4)! \):
\[ 24(n-4)! = 4(n-2)(n-3)(n-4)! \]

6. **Cancel \( (n-4)! \) from both sides** (assuming \( n \geq 4 \)):
\[ 24 = 4(n-2)(n-3) \]

7. **Divide both sides by 4**:
\[ 6 = (n-2)(n-3) \]

8. **Expand the equation**:
\[ n^2 - 5n + 6 = 0 \]

9. **Factor the quadratic equation**:
\[ (n-2)(n-3) = 0 \]

10. **Find the values of \( n \)**:
\[ n - 2 = 0 \quad \Rightarrow \quad n = 2 \]
\[ n - 3 = 0 \quad \Rightarrow \quad n = 3 \]

11. **Check for valid values of \( n \)**:
Since we derived the equation under the assumption \( n \geq 4 \), we discard \( n = 2 \) and \( n = 3 \).

12. **Conclusion**:
The only valid solution is \( n = 5 \).

### Final Answer:
\[ n = 5 \]