If `"^(n)C_(n-3) = 120`, find n.

Answer

To solve the equation \( \binom{n}{n-3} = 120 \), we will follow these steps:

### Step 1: Write the formula for combinations
The combination formula is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
In our case, \( r = n - 3 \). Therefore, we can rewrite the equation as:
\[ \binom{n}{n-3} = \frac{n!}{(n-3)! \cdot 3!} \]

### Step 2: Substitute the value into the equation
Substituting into the equation, we have:
\[ \frac{n!}{(n-3)! \cdot 3!} = 120 \]

### Step 3: Simplify the equation
We know that \( 3! = 6 \). Thus, the equation becomes:
\[ \frac{n!}{(n-3)! \cdot 6} = 120 \]
Multiplying both sides by 6 gives:
\[ \frac{n!}{(n-3)!} = 720 \]

### Step 4: Expand \( n! \)
We can express \( n! \) as:
\[ n! = n \times (n-1) \times (n-2) \times (n-3)! \]
Substituting this into the equation gives:
\[ \frac{n \times (n-1) \times (n-2) \times (n-3)!}{(n-3)!} = 720 \]
The \( (n-3)! \) cancels out:
\[ n \times (n-1) \times (n-2) = 720 \]

### Step 5: Solve the equation
We need to find integers \( n \) such that:
\[ n(n-1)(n-2) = 720 \]
To find suitable values for \( n \), we can test integer values:

- For \( n = 10 \):
\[ 10 \times 9 \times 8 = 720 \]
This works!

### Conclusion
Thus, the value of \( n \) is:
\[ \boxed{10} \]