Find `n if (^(2n)C_2)/(^(n)C_2)= 12:1`

Answer

To solve the problem, we need to find the value of \( n \) such that

\[ \frac{{^{2n}C_2}}{{^{n}C_2}} = \frac{12}{1} \]

### Step-by-Step Solution:

1. **Write the Combination Formula**:
The combination formula is given by:
\[ ^nC_r = \frac{n!}{r!(n-r)!} \]
Therefore, we can express \( ^{2n}C_2 \) and \( ^nC_2 \) using this formula.

2. **Express \( ^{2n}C_2 \) and \( ^nC_2 \)**:
\[ ^{2n}C_2 = \frac{(2n)!}{2!(2n-2)!} \]
\[ ^nC_2 = \frac{n!}{2!(n-2)!} \]

3. **Substitute into the Ratio**:
Substitute these expressions into the ratio:
\[ \frac{{^{2n}C_2}}{{^{n}C_2}} = \frac{\frac{(2n)!}{2!(2n-2)!}}{\frac{n!}{2!(n-2)!}} = \frac{(2n)! \cdot (n-2)!}{(2n-2)! \cdot n!} \]

4. **Simplify the Ratio**:
The \( 2! \) cancels out, so we have:
\[ \frac{(2n)! \cdot (n-2)!}{(2n-2)! \cdot n!} = 12 \]

5. **Use Factorial Relationships**:
We can express \( (2n)! \) as \( (2n)(2n-1)(2n-2)! \) and \( n! \) as \( n(n-1)(n-2)! \):
\[ \frac{(2n)(2n-1)(2n-2)! \cdot (n-2)!}{(2n-2)! \cdot n(n-1)(n-2)!} = 12 \]
The \( (2n-2)! \) and \( (n-2)! \) cancel out:
\[ \frac{(2n)(2n-1)}{n(n-1)} = 12 \]

6. **Cross Multiply**:
Cross multiplying gives:
\[ (2n)(2n-1) = 12n(n-1) \]

7. **Expand Both Sides**:
Expanding both sides results in:
\[ 4n^2 - 2n = 12n^2 - 12n \]

8. **Rearranging the Equation**:
Rearranging gives:
\[ 4n^2 - 12n^2 + 12n - 2n = 0 \]
\[ -8n^2 + 10n = 0 \]

9. **Factor Out Common Terms**:
Factoring out \( 2n \):
\[ 2n(-4n + 5) = 0 \]

10. **Solve for \( n \)**:
This gives us two solutions:
\[ 2n = 0 \quad \text{or} \quad -4n + 5 = 0 \]
From \( 2n = 0 \), we get \( n = 0 \).
From \( -4n + 5 = 0 \), we get \( n = \frac{5}{4} = 1.25 \).

11. **Conclusion**:
Since \( n \) must be a natural number, the only valid solution is \( n = 0 \).