`"^mC_1=^nC_2`, then -

Options

A. 2m=n
B. 2m=n(n+1)
C. 2m=n(n-1)
D. 2n=m(m-1)

Answer

To solve the equation \( ^mC_1 = ^nC_2 \), we will use the formula for combinations.

### Step-by-Step Solution:

1. **Write the Combination Formulas**:
The formula for combinations is given by:
\[ ^rC_k = \frac{r!}{(r-k)! \cdot k!} \]
For \( ^mC_1 \) and \( ^nC_2 \), we can write:
\[ ^mC_1 = \frac{m!}{(m-1)! \cdot 1!} = m \]
\[ ^nC_2 = \frac{n!}{(n-2)! \cdot 2!} = \frac{n(n-1)}{2} \]

2. **Set the Equations Equal**:
Since \( ^mC_1 = ^nC_2 \), we have:
\[ m = \frac{n(n-1)}{2} \]

3. **Multiply Both Sides by 2**:
To eliminate the fraction, multiply both sides by 2:
\[ 2m = n(n-1) \]

4. **Rearrange the Equation**:
This can be rearranged to:
\[ n^2 - n - 2m = 0 \]

5. **Identify the Quadratic Equation**:
The equation \( n^2 - n - 2m = 0 \) is a quadratic equation in terms of \( n \).

6. **Use the Quadratic Formula**:
The solutions for \( n \) can be found using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -1, c = -2m \):
\[ n = \frac{1 \pm \sqrt{1 + 8m}}{2} \]

### Conclusion:
Thus, we have derived the relationship between \( m \) and \( n \) as:
\[ 2m = n(n-1) \]