The smallest value of x satisfying the inequality `""^(10)C_(x-1)gt2.""^(10)C_(x)` is.

Options

A. 7
B. 10
C. 9
D. 8

Answer

To solve the inequality \( \binom{10}{x-1} > 2 \cdot \binom{10}{x} \), we will follow these steps:

### Step 1: Write the inequality using the definition of combinations
The combination formula is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Applying this to our inequality, we have:
\[ \binom{10}{x-1} = \frac{10!}{(x-1)!(10-(x-1))!} = \frac{10!}{(x-1)!(11-x)!} \]
\[ \binom{10}{x} = \frac{10!}{x!(10-x)!} \]
Thus, our inequality becomes:
\[ \frac{10!}{(x-1)!(11-x)!} > 2 \cdot \frac{10!}{x!(10-x)!} \]

### Step 2: Simplify the inequality
We can cancel \( 10! \) from both sides:
\[ \frac{1}{(x-1)!(11-x)!} > 2 \cdot \frac{1}{x!(10-x)!} \]
This simplifies to:
\[ \frac{1}{(11-x)!} > 2 \cdot \frac{1}{x \cdot (10-x)!} \]

### Step 3: Rearranging the inequality
Multiplying both sides by \( (x-1)! \cdot (10-x)! \) (which is positive), we get:
\[ (10-x)! > 2 \cdot \frac{(11-x)!}{x} \]
This can be rewritten as:
\[ (10-x)! > \frac{2 \cdot (11-x)(10-x)!}{x} \]
Cancelling \( (10-x)! \) from both sides (for \( x < 10 \)):
\[ 1 > \frac{2(11-x)}{x} \]

### Step 4: Cross-multiplying
Cross-multiplying gives us:
\[ x > 2(11-x) \]
Expanding this, we get:
\[ x > 22 - 2x \]

### Step 5: Solving for x
Adding \( 2x \) to both sides:
\[ 3x > 22 \]
Dividing both sides by 3:
\[ x > \frac{22}{3} \approx 7.33 \]

### Step 6: Finding the smallest integer value
Since \( x \) must be an integer, the smallest integer satisfying this inequality is:
\[ x \geq 8 \]

### Conclusion
Thus, the smallest value of \( x \) satisfying the inequality \( \binom{10}{x-1} > 2 \cdot \binom{10}{x} \) is:
\[ \boxed{8} \]