If `({:(6),(x):})=({:(4),(x))`, then x=

Options

A. 0
B. 1
C. 4
D. 5

Explanation

To solve the equation \( \binom{6}{x} = \binom{4}{x} \), we can use the properties of combinations.

### Step-by-Step Solution:

1. **Understanding Combinations**: Recall that \( \binom{n}{r} \) (read as "n choose r") is the number of ways to choose \( r \) elements from a set of \( n \) elements. The formula is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

2. **Setting Up the Equation**: We start with the equation:
\[ \binom{6}{x} = \binom{4}{x} \]

3. **Applying the Definition of Combinations**:
\[ \frac{6!}{x!(6-x)!} = \frac{4!}{x!(4-x)!} \]

4. **Cancelling Common Terms**: Since \( x! \) appears in both sides, we can cancel it out (assuming \( x \) is not negative):
\[ \frac{6!}{(6-x)!} = \frac{4!}{(4-x)!} \]

5. **Cross-Multiplying**: We can cross-multiply to eliminate the fractions:
\[ 6! \cdot (4-x)! = 4! \cdot (6-x)! \]

6. **Substituting Factorials**: We know that \( 6! = 6 \times 5 \times 4! \), so we can substitute:
\[ 6 \times 5 \times 4! \cdot (4-x)! = 4! \cdot (6-x)! \]

7. **Dividing by \( 4! \)**: We can divide both sides by \( 4! \):
\[ 6 \times 5 \cdot (4-x)! = (6-x)! \]

8. **Expanding Factorials**: The factorial \( (6-x)! \) can be expressed as:
\[ (6-x)(5-x)(4-x)! \]
So, we rewrite the equation:
\[ 6 \times 5 \cdot (4-x)! = (6-x)(5-x)(4-x)! \]

9. **Cancelling \( (4-x)! \)**: Assuming \( 4-x \neq 0 \), we can cancel \( (4-x)! \):
\[ 6 \times 5 = (6-x)(5-x) \]

10. **Expanding the Right Side**:
\[ 30 = (6-x)(5-x) \]

11. **Expanding the Product**:
\[ 30 = 30 - 11x + x^2 \]

12. **Rearranging the Equation**:
\[ x^2 - 11x = 0 \]

13. **Factoring**:
\[ x(x - 11) = 0 \]

14. **Finding the Solutions**: This gives us two solutions:
\[ x = 0 \quad \text{or} \quad x = 11 \]

15. **Validating the Solutions**: Since \( x \) must be less than or equal to 4 in \( \binom{4}{x} \), the only valid solution is:
\[ x = 0 \]

### Final Answer:
Thus, the value of \( x \) is \( 0 \).