The solution set `"^10C_(x-1)>2`.`"^10C_x` is

Options

A. {1,2,3}
B. {4,5,6}
C. {8,9,10}
D. {9,10,11}

Answer

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

### Step 1: Rewrite the Combinations
Using the combination formula, we can express \( \binom{10}{x-1} \) and \( \binom{10}{x} \) as follows:

\[ \binom{10}{x-1} = \frac{10!}{(x-1)!(10-(x-1))!} = \frac{10!}{(x-1)!(11-x)!} \]

\[ \binom{10}{x} = \frac{10!}{x!(10-x)!} \]

### Step 2: Substitute into the Inequality
Now we substitute these expressions into the inequality:

\[ \frac{10!}{(x-1)!(11-x)!} > 2 \cdot \frac{10!}{x!(10-x)!} \]

### Step 3: Cancel \( 10! \) from Both Sides
We can cancel \( 10! \) from both sides:

\[ \frac{1}{(x-1)!(11-x)!} > \frac{2}{x!(10-x)!} \]

### Step 4: Rewrite the Right Side
The right side can be rewritten as:

\[ \frac{2}{x \cdot (x-1)!(10-x)!} \]

### Step 5: Cross Multiply
Cross multiplying gives us:

\[ (10-x)! > 2 \cdot (11-x)! \]

### Step 6: Simplify the Inequality
We can simplify \( (11-x)! \) as \( (11-x) \cdot (10-x)! \):

\[ (10-x)! > 2 \cdot (11-x) \cdot (10-x)! \]

### Step 7: Cancel \( (10-x)! \)
Assuming \( (10-x)! \neq 0 \), we can cancel it from both sides:

\[ 1 > 2 \cdot (11-x) \]

### Step 8: Solve for \( x \)
Rearranging gives:

\[ 1 > 22 - 2x \]

\[ 2x > 21 \]

\[ x > \frac{21}{2} = 10.5 \]

### Step 9: Determine Integer Values
Since \( x \) must be an integer, the smallest integer greater than \( 10.5 \) is \( 11 \).

### Step 10: Check Validity of \( x \)
Now we need to check the constraints for \( x \):

1. From \( \binom{10}{x-1} \), we need \( x-1 \leq 10 \) which gives \( x \leq 11 \).
2. From \( \binom{10}{x} \), we need \( x \leq 10 \).

Thus, the only integer value satisfying both conditions is \( x = 11 \).

### Conclusion
The solution set for the inequality \( \binom{10}{x-1} > 2 \cdot \binom{10}{x} \) is:

\[ \{ 11 \} \]