If `""^(n)C_(10)= ""^(n)C_(9)`, find `""^(n)C_(19)`.

Answer

To solve the problem where \( \binom{n}{10} = \binom{n}{9} \) and find \( \binom{n}{19} \), we can follow these steps:

### Step 1: Set up the equation
Given the equation \( \binom{n}{10} = \binom{n}{9} \), we can use the definition of combinations:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Thus, we have:
\[ \binom{n}{10} = \frac{n!}{10!(n-10)!} \quad \text{and} \quad \binom{n}{9} = \frac{n!}{9!(n-9)!} \]

### Step 2: Write the equation using factorials
Setting the two expressions equal gives:
\[ \frac{n!}{10!(n-10)!} = \frac{n!}{9!(n-9)!} \]

### Step 3: Cancel out \( n! \)
Since \( n! \) is common on both sides, we can cancel it out (assuming \( n! \neq 0 \)):
\[ \frac{1}{10!(n-10)!} = \frac{1}{9!(n-9)!} \]

### Step 4: Cross-multiply
Cross-multiplying leads to:
\[ 9!(n-9)! = 10!(n-10)! \]

### Step 5: Simplify the factorials
We can express \( 10! \) as \( 10 \times 9! \):
\[ 9!(n-9)! = 10 \times 9!(n-10)! \]
Dividing both sides by \( 9! \) (assuming \( 9! \neq 0 \)):
\[ (n-9)! = 10(n-10)! \]

### Step 6: Rewrite \( (n-9)! \)
We can express \( (n-9)! \) as:
\[ (n-9)(n-10)! = 10(n-10)! \]

### Step 7: Cancel \( (n-10)! \)
Assuming \( (n-10)! \neq 0 \), we can cancel it:
\[ n-9 = 10 \]

### Step 8: Solve for \( n \)
Solving for \( n \):
\[ n = 19 \]

### Step 9: Find \( \binom{n}{19} \)
Now, we need to find \( \binom{n}{19} \):
\[ \binom{19}{19} = 1 \]

### Final Answer
Thus, the value of \( \binom{n}{19} \) is:
\[ \boxed{1} \]