Determine `n,` if `""^(2n) C_(2): ""^(n) C_(2) = 12: 1`.

Options

A. `n=7`
B. `n=3`
C. `n=5`
D. `n=2`

Answer

To solve the problem, we need to determine \( n \) such that the ratio of \( \binom{2n}{2} \) to \( \binom{n}{2} \) is equal to \( 12:1 \).

### Step-by-Step Solution:

1. **Write the formula for combinations**:
The formula for combinations is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

2. **Calculate \( \binom{2n}{2} \)**:
Using the formula, we have:
\[ \binom{2n}{2} = \frac{(2n)!}{2!(2n-2)!} \]
Simplifying this, we can express it as:
\[ \binom{2n}{2} = \frac{2n(2n-1)}{2} = n(2n-1) \]

3. **Calculate \( \binom{n}{2} \)**:
Similarly, we calculate:
\[ \binom{n}{2} = \frac{n!}{2!(n-2)!} = \frac{n(n-1)}{2} \]

4. **Set up the ratio**:
According to the problem, we have:
\[ \frac{\binom{2n}{2}}{\binom{n}{2}} = \frac{n(2n-1)}{\frac{n(n-1)}{2}} = \frac{2n(2n-1)}{n(n-1)} \]
This should equal \( 12 \):
\[ \frac{2n(2n-1)}{n(n-1)} = 12 \]

5. **Simplify the equation**:
Cancel \( n \) from numerator and denominator (assuming \( n \neq 0 \)):
\[ \frac{2(2n-1)}{n-1} = 12 \]

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

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

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

9. **Solve for \( n \)**:
Dividing both sides by -8:
\[ n = \frac{10}{8} = \frac{5}{4} \]

### Final Answer:
Thus, the value of \( n \) is \( \frac{5}{4} \).