If ` C (2n,3) : C ( n,2)=12 : 1 ` find n,

Answer

To solve the problem \( \frac{C(2n, 3)}{C(n, 2)} = \frac{12}{1} \), we will follow these steps:

### Step 1: Write the combinations in terms of factorials
The combination formula is given by:
\[ C(n, r) = \frac{n!}{r!(n-r)!} \]
Thus, we can express \( C(2n, 3) \) and \( C(n, 2) \) as follows:
\[ C(2n, 3) = \frac{(2n)!}{3!(2n-3)!} \]
\[ C(n, 2) = \frac{n!}{2!(n-2)!} \]

### Step 2: Set up the equation
Substituting these into the equation gives:
\[ \frac{(2n)!}{3!(2n-3)!} \div \frac{n!}{2!(n-2)!} = 12 \]
This can be rewritten as:
\[ \frac{(2n)! \cdot 2!(n-2)!}{3!(2n-3)! \cdot n!} = 12 \]

### Step 3: Simplify the equation
Now, we can simplify the left-hand side:
\[ \frac{(2n)! \cdot 2(n-2)!}{6(2n-3)! \cdot n!} = 12 \]
This simplifies to:
\[ \frac{(2n)(2n-1)(2n-2)(2n-3)! \cdot 2(n-2)!}{6(2n-3)! \cdot n!} = 12 \]
Canceling \( (2n-3)! \) from both sides gives:
\[ \frac{(2n)(2n-1)(2n-2) \cdot 2(n-2)!}{6 \cdot n!} = 12 \]

### Step 4: Further simplification
Now, we can cancel \( 2 \) from the numerator and the denominator:
\[ \frac{(2n)(2n-1)(2n-2)(n-2)!}{3 \cdot n!} = 12 \]

### Step 5: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ (2n)(2n-1)(2n-2)(n-2)! = 36 \cdot n! \]

### Step 6: Substitute \( n! \)
We know that \( n! = n(n-1)(n-2)! \), so we can substitute this into the equation:
\[ (2n)(2n-1)(2n-2)(n-2)! = 36 \cdot n(n-1)(n-2)! \]
Cancelling \( (n-2)! \) from both sides gives:
\[ (2n)(2n-1)(2n-2) = 36n(n-1) \]

### Step 7: Expand and simplify
Expanding both sides:
\[ 8n^3 - 12n^2 + 4n = 36n^2 - 36n \]
Rearranging gives:
\[ 8n^3 - 48n^2 + 40n = 0 \]

### Step 8: Factor out common terms
Factoring out \( 8n \):
\[ 8n(n^2 - 6n + 5) = 0 \]
This gives us:
\[ n = 0 \quad \text{or} \quad n^2 - 6n + 5 = 0 \]

### Step 9: Solve the quadratic equation
Factoring the quadratic:
\[ (n-1)(n-5) = 0 \]
Thus, \( n = 1 \) or \( n = 5 \).

### Step 10: Conclusion
Since \( n \) must be a positive integer, we have:
\[ n = 5 \]