If `""^(k^2-k) C_2= "^(k^2-k) C_4` then k=

Options

A. 2
B. 3
C. 4
D. none of these

Answer

To solve the equation \( \binom{k^2 - k}{2} = \binom{k^2 - k}{4} \), we can use the property of combinations that states \( \binom{n}{r} = \binom{n}{n-r} \).

### Step-by-Step Solution:

1. **Write the Combinations Using the Property**:
\[ \binom{k^2 - k}{4} = \binom{k^2 - k}{(k^2 - k) - 4} \]
This means we can rewrite the equation as:
\[ \binom{k^2 - k}{2} = \binom{k^2 - k}{(k^2 - k) - 4} \]

2. **Set the Arguments Equal**:
Since the left-hand side is equal to the right-hand side, we can equate the second argument of the right-hand side:
\[ 2 = (k^2 - k) - 4 \]

3. **Rearrange the Equation**:
Rearranging gives:
\[ k^2 - k - 6 = 0 \]

4. **Factor the Quadratic Equation**:
We can factor the quadratic:
\[ (k - 3)(k + 2) = 0 \]

5. **Find the Values of k**:
Setting each factor to zero gives:
\[ k - 3 = 0 \quad \Rightarrow \quad k = 3 \]
\[ k + 2 = 0 \quad \Rightarrow \quad k = -2 \]

6. **Check the Validity of the Solutions**:
Since \( k \) must be a non-negative integer in the context of combinations, we discard \( k = -2 \). Thus, the only valid solution is:
\[ k = 3 \]

### Final Answer:
\[ k = 3 \]