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 we need to find \( ^nC_{17} \) given that \( ^nC_{9} = ^nC_{8} \), we can follow these steps:

### Step 1: Set up the equation using the definition of combinations
We know that the combination formula is given by:
\[ ^nC_r = \frac{n!}{r!(n-r)!} \]
Using this, we can express \( ^nC_{9} \) and \( ^nC_{8} \):
\[ ^nC_{9} = \frac{n!}{9!(n-9)!} \]
\[ ^nC_{8} = \frac{n!}{8!(n-8)!} \]
Given that \( ^nC_{9} = ^nC_{8} \), we can set these two equations equal to each other:
\[ \frac{n!}{9!(n-9)!} = \frac{n!}{8!(n-8)!} \]

### Step 2: Cancel \( n! \) from both sides
Since \( n! \) appears in both the numerator, we can cancel it:
\[ \frac{1}{9!(n-9)!} = \frac{1}{8!(n-8)!} \]

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

### Step 4: Substitute \( 9! \) and simplify
Recall that \( 9! = 9 \times 8! \):
\[ 8!(n-8)! = 9 \times 8!(n-9)! \]
Now, we can cancel \( 8! \) from both sides:
\[ (n-8)! = 9(n-9)! \]

### Step 5: Use the factorial property
Using the property \( (n-8)! = (n-8)(n-9)! \), we can substitute:
\[ (n-8)(n-9)! = 9(n-9)! \]
Now, we can cancel \( (n-9)! \) from both sides (assuming \( n \neq 9 \)):
\[ n-8 = 9 \]

### Step 6: Solve for \( n \)
Adding 8 to both sides gives:
\[ n = 17 \]

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

### Final Answer
Thus, \( ^nC_{17} = 1 \).

---