The number of solutions of the inequation `""^(10)C_(x-1)gt3.""^(10)C_(x)` is

Options

A. 0
B. 1
C. 2
D. 9

Explanation

To solve the inequality \( \binom{10}{x-1} > 3 \binom{10}{x} \), we will follow these steps:

### Step 1: Understand the Binomial Coefficient Inequality
The binomial coefficient \( \binom{n}{k} \) represents the number of ways to choose \( k \) elements from \( n \) elements. The inequality \( \binom{10}{x-1} > 3 \binom{10}{x} \) can be rewritten using the properties of binomial coefficients.

### Step 2: Rewrite the Inequality
Using the property of binomial coefficients, we know that:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
Thus, we can express the inequality as:
\[ \frac{10!}{(x-1)!(10-(x-1))!} > 3 \cdot \frac{10!}{x!(10-x)!} \]
This simplifies to:
\[ \frac{1}{(x-1)!(11-x)!} > \frac{3}{x!(10-x)!} \]

### Step 3: Simplify the Inequality
We can cancel \( 10! \) from both sides:
\[ \frac{x}{11-x} > 3 \]
Cross-multiplying gives:
\[ x > 3(11 - x) \]
This simplifies to:
\[ x > 33 - 3x \]

### Step 4: Solve for \( x \)
Rearranging the inequality:
\[ x + 3x > 33 \]
\[ 4x > 33 \]
\[ x > \frac{33}{4} = 8.25 \]

### Step 5: Determine the Integer Solutions
Since \( x \) must be an integer, the smallest integer satisfying \( x > 8.25 \) is \( x = 9 \). However, we also need to consider the upper limit since \( x \) must be less than or equal to 10 (the maximum value for \( \binom{10}{k} \)).

### Step 6: Identify Valid Integer Values
Thus, the possible integer values for \( x \) are:
- \( x = 9 \)
- \( x = 10 \)

### Conclusion
The number of solutions to the inequality \( \binom{10}{x-1} > 3 \binom{10}{x} \) is **2** (for \( x = 9 \) and \( x = 10 \)).