The smallest value of r satisfying the inequality `""^10 C_(r-1) gt 2"^10 C_r` is

Options

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

Answer

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

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

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

### Step 3: Rewrite the left-hand side
Notice that \(r! = r \cdot (r-1)!\), so we can rewrite the left-hand side:
\[ \frac{(10-r)!}{(r-1)!(11-r)!} = \frac{(10-r)!}{(11-r) \cdot (10-r)!} = \frac{1}{11-r} \]
Thus, the inequality becomes:
\[ \frac{1}{11-r} > 2 \]

### Step 4: Solve the inequality
To solve \( \frac{1}{11-r} > 2 \), we can take the reciprocal (and reverse the inequality):
\[ 11 - r < \frac{1}{2} \]
Rearranging gives:
\[ r > 11 - \frac{1}{2} = 10.5 \]

### Step 5: Determine the smallest integer value of \(r\)
Since \(r\) must be an integer, the smallest integer satisfying \(r > 10.5\) is:
\[ r = 11 \]

### Step 6: Verify the solution
We can check if \(r = 11\) satisfies the original inequality:
- For \(r = 11\), \( \binom{10}{10} = 1 \) and \( \binom{10}{11} = 0 \), thus \( 1 > 2 \cdot 0\) holds true.

### Conclusion
The smallest value of \(r\) satisfying the inequality is:
\[ \boxed{11} \]