Find the value of n if `(n)/(11!) = (1)/(9!) +(1)/(10!)`.

Answer

To solve the equation \(\frac{n}{11!} = \frac{1}{9!} + \frac{1}{10!}\), we will follow these steps:

### Step 1: Rewrite the Right Side
We start with the equation:
\[ \frac{n}{11!} = \frac{1}{9!} + \frac{1}{10!} \]
We need to find a common denominator for the right side. The common denominator of \(9!\) and \(10!\) is \(10!\). Thus, we can rewrite the right side as:
\[ \frac{1}{9!} = \frac{10}{10!} \]
So,
\[ \frac{1}{9!} + \frac{1}{10!} = \frac{10}{10!} + \frac{1}{10!} = \frac{10 + 1}{10!} = \frac{11}{10!} \]

### Step 2: Substitute Back into the Equation
Now we substitute this back into the equation:
\[ \frac{n}{11!} = \frac{11}{10!} \]

### Step 3: Cross-Multiply
Next, we cross-multiply to eliminate the fractions:
\[ n \cdot 10! = 11 \cdot 11! \]

### Step 4: Simplify \(11!\)
Recall that \(11! = 11 \times 10!\). Therefore, we can substitute this into our equation:
\[ n \cdot 10! = 11 \cdot (11 \times 10!) \]
This simplifies to:
\[ n \cdot 10! = 121 \cdot 10! \]

### Step 5: Divide Both Sides by \(10!\)
Since \(10!\) is common on both sides, we can divide both sides by \(10!\):
\[ n = 121 \]

### Final Answer
Thus, the value of \(n\) is:
\[ \boxed{121} \]