If `""^(n)C_(8)= ""^(n)C_(9)`, find the value of n.

Answer

To solve the equation \( \binom{n}{8} = \binom{n}{9} \), we can follow these steps:

### Step 1: Write the combinations in terms of factorials
The formula for combinations is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Using this, we can express \( \binom{n}{8} \) and \( \binom{n}{9} \):
\[ \binom{n}{8} = \frac{n!}{8!(n-8)!} \]
\[ \binom{n}{9} = \frac{n!}{9!(n-9)!} \]

### Step 2: Set the two expressions equal to each other
From the problem, we have:
\[ \frac{n!}{8!(n-8)!} = \frac{n!}{9!(n-9)!} \]

### Step 3: Cancel \( n! \) from both sides
Since \( n! \) is common in both expressions, we can cancel it out:
\[ \frac{1}{8!(n-8)!} = \frac{1}{9!(n-9)!} \]

### Step 4: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ 9!(n-9)! = 8!(n-8)! \]

### Step 5: Rewrite \( 9! \) and \( (n-8)! \)
We can rewrite \( 9! \) as \( 9 \times 8! \):
\[ 9 \times 8!(n-9)! = 8!(n-8)! \]

### Step 6: Cancel \( 8! \) from both sides
Now, we can cancel \( 8! \):
\[ 9(n-9)! = (n-8)! \]

### Step 7: Rewrite \( (n-8)! \) in terms of \( (n-9)! \)
We can express \( (n-8)! \) as:
\[ (n-8)! = (n-8)(n-9)! \]
Substituting this into the equation gives:
\[ 9(n-9)! = (n-8)(n-9)! \]

### Step 8: Cancel \( (n-9)! \) from both sides
Assuming \( n-9 \neq 0 \), we can cancel \( (n-9)! \):
\[ 9 = n - 8 \]

### Step 9: Solve for \( n \)
Now, we can solve for \( n \):
\[ n = 9 + 8 = 17 \]

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

---