If `^n C_9=^n C_8,`find `^nC_17`

Options

A. 17
B. 1
C. 17!
D. None of these

Explanation

To solve the problem where \( ^nC_9 = ^nC_8 \), we will follow these steps:

### Step 1: Understand the given equation
We know from the properties of combinations that:
\[ ^nC_r = \frac{n!}{r!(n-r)!} \]
Given \( ^nC_9 = ^nC_8 \), we can express this as:
\[ \frac{n!}{9!(n-9)!} = \frac{n!}{8!(n-8)!} \]

### Step 2: Simplify the equation
Since \( n! \) appears in both sides, we can cancel it out (assuming \( n \geq 9 \)):
\[ \frac{1}{9!(n-9)!} = \frac{1}{8!(n-8)!} \]

### Step 3: Cross-multiply
Cross-multiplying gives us:
\[ 8!(n-8)! = 9!(n-9)! \]

### Step 4: Substitute \( 9! \) and simplify
We know that \( 9! = 9 \times 8! \), so we can substitute:
\[ 8!(n-8)! = 9 \times 8!(n-9)! \]
Now, we can cancel \( 8! \) from both sides (again assuming \( n \geq 8 \)):
\[ (n-8)! = 9(n-9)! \]

### Step 5: Simplify further
Using the fact that \( (n-8)! = (n-8)(n-9)! \), we can rewrite the equation:
\[ (n-8)(n-9)! = 9(n-9)! \]
Now, we can cancel \( (n-9)! \) from both sides (assuming \( n \geq 9 \)):
\[ n - 8 = 9 \]

### Step 6: Solve for \( n \)
Solving the equation gives:
\[ n = 17 \]

### Step 7: Find \( ^nC_{17} \)
Now that we have \( n = 17 \), we need to find \( ^{17}C_{17} \):
\[ ^{17}C_{17} = \frac{17!}{17! \cdot 0!} = \frac{17!}{17! \cdot 1} = 1 \]

### Final Answer
Thus, the value of \( ^{n}C_{17} \) is:
\[ \boxed{1} \]