12. ENG: Thic value of \( x \geq 0 \) for which \( C_{x}^{x}=3 \) is: FRE: La valeur de \( x \geq 0 \) pour laquelle \( C_{x}^{2}=3 \) est:

To solve the equation \( C_{x}^{x} = 3 \), we first need to understand what \( C_{x}^{x} \) represents. The notation \( C_{n}^{k} \) typically refers to the binomial coefficient, which is defined as: \[ C_{n}^{k} = \frac{n!}{k!(n-k)!} \] In this case, we have \( C_{x}^{x} \), which simplifies to: \[ C_{x}^{x} = \frac{x!}{x! \cdot 0!} = \frac{x!}{x! \cdot 1} = 1 \] This means that \( C_{x}^{x} \) is always equal to 1 for any non-negative integer \( x \). Therefore, there is no value of \( x \) such that \( C_{x}^{x} = 3 \). Now, for the French part of the question, it seems there is a typo in the notation. The correct interpretation should likely be \( C_{x}^{2} = 3 \). To solve \( C_{x}^{2} = 3 \): \[ C_{x}^{2} = \frac{x!}{2!(x-2)!} = \frac{x(x-1)}{2} \] Setting this equal to 3 gives: \[ \frac{x(x-1)}{2} = 3 \] Multiplying both sides by 2: \[ x(x-1) = 6 \] Rearranging gives: \[ x^2 - x - 6 = 0 \] Factoring the quadratic: \[ (x - 3)(x + 2) = 0 \] This gives us the solutions: \[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] \[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \] Since we are looking for \( x \geq 0 \), the only valid solution is: \[ x = 3 \] Thus, the value of \( x \geq 0 \) for which \( C_{x}^{2} = 3 \) is \( x = 3 \).