The number of solution of the in equation `""^10C_(x-1) gt 3. ""^10C_x ` is :

Options

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

Explanation

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

### Step 1: Rewrite the Binomial Coefficients
The binomial coefficient \( \binom{n}{r} \) is defined as:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Thus, we can express \( \binom{10}{x-1} \) and \( \binom{10}{x} \) as:
\[ \binom{10}{x-1} = \frac{10!}{(x-1)!(10-(x-1))!} = \frac{10!}{(x-1)!(11-x)!} \]
\[ \binom{10}{x} = \frac{10!}{x!(10-x)!} = \frac{10!}{x!(10-x)!} \]

### Step 2: Substitute into the Inequality
Substituting these into the inequality gives:
\[ \frac{10!}{(x-1)!(11-x)!} > 3 \cdot \frac{10!}{x!(10-x)!} \]

### Step 3: Cancel Common Terms
Since \( 10! \) appears in both sides, we can cancel it:
\[ \frac{1}{(x-1)!(11-x)!} > 3 \cdot \frac{1}{x!(10-x)!} \]

### Step 4: Rearranging the Inequality
Rearranging gives:
\[ \frac{1}{(x-1)!(11-x)!} > \frac{3}{x!(10-x)!} \]
This can be rewritten as:
\[ x!(10-x)! > 3(x-1)!(11-x)! \]

### Step 5: Simplifying Further
We can express \( x! \) as \( x \cdot (x-1)! \) and \( (10-x)! \) as \( (10-x)(9-x)! \):
\[ x \cdot (x-1)! \cdot (10-x)(9-x)! > 3(x-1)!(11-x)! \]
Cancelling \( (x-1)! \) from both sides gives:
\[ x(10-x)(9-x)! > 3(11-x)! \]

### Step 6: Further Simplification
Now, we can express \( (11-x)! \) as \( (11-x)(10-x)! \):
\[ x(10-x)(9-x)! > 3(11-x)(10-x)! \]
Cancelling \( (10-x)! \) gives:
\[ x(9-x)! > 3(11-x) \]

### Step 7: Analyzing the Inequality
Now, we need to find the values of \( x \) that satisfy this inequality.

### Step 8: Finding the Range of \( x \)
Since \( x \) must be a non-negative integer and \( x \) can take values from 0 to 10, we can test integer values of \( x \):

1. For \( x = 0 \): \( 0 > 33 \) (false)
2. For \( x = 1 \): \( 1 > 30 \) (false)
3. For \( x = 2 \): \( 2 > 27 \) (false)
4. For \( x = 3 \): \( 3 > 24 \) (false)
5. For \( x = 4 \): \( 4 > 21 \) (false)
6. For \( x = 5 \): \( 5 > 18 \) (false)
7. For \( x = 6 \): \( 6 > 15 \) (false)
8. For \( x = 7 \): \( 7 > 12 \) (false)
9. For \( x = 8 \): \( 8 > 9 \) (false)
10. For \( x = 9 \): \( 9 > 6 \) (true)
11. For \( x = 10 \): \( 10 > 3 \) (true)

### Conclusion
The valid integer solutions for \( x \) are \( x = 9 \) and \( x = 10 \). Therefore, the number of solutions is **2**.