We are given the equation involving binomial coefficients:

\[
\binom{n}{n-3} = 3 \cdot \binom{n}{2}
\]

### Step 1: Simplify the binomial coefficients
Using the binomial coefficient formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), we can express both sides of the equation.

#### Left-hand side:
\[
\binom{n}{n-3} = \binom{n}{3} = \frac{n!}{3!(n-3)!}
\]

#### Right-hand side:
\[
\binom{n}{2} = \frac{n!}{2!(n-2)!} = \frac{n(n-1)}{2}
\]

Thus, the equation becomes:
\[
\frac{n!}{3!(n-3)!} = 3 \cdot \frac{n(n-1)}{2}
\]

### Step 2: Simplify the factorials
We know that \(3! = 6\), so the equation is:
\[
\frac{n(n-1)(n-2)}{6} = 3 \cdot \frac{n(n-1)}{2}
\]

### Step 3: Simplify both sides
Now, simplify the right-hand side:
\[
3 \cdot \frac{n(n-1)}{2} = \frac{3n(n-1)}{2}
\]

Thus, the equation becomes:
\[
\frac{n(n-1)(n-2)}{6} = \frac{3n(n-1)}{2}
\]

### Step 4: Eliminate common terms
Cancel \(n(n-1)\) from both sides (assuming \(n \neq 0\) and \(n \neq 1\)):
\[
\frac{n-2}{6} = \frac{3}{2}
\]

### Step 5: Solve for \(n\)
Multiply both sides by 6:
\[
n - 2 = 9
\]

Thus, \(n = 11\).

### Conclusion:
The value of \(n\) that satisfies the equation is \(n = 11\).

Would you like more details or have any questions?

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Here are some related questions:
1. How would the equation change if the binomial coefficient was \(\binom{n}{n-4}\)?
2. Can you explain the relationship between different binomial coefficient identities?
3. What happens if we change the right-hand side constant to \(k\cdot\binom{n}{2}\)?
4. How does the binomial expansion apply to combinatorial identities?
5. Can you solve a similar equation for a different binomial form?

**Tip:** When simplifying equations involving factorials, look for common terms to cancel early.

Learn how to solve the binomial coefficient equation binom(n,n-3) = 3 * binom(n,2) with step-by-step simplifications and factorial manipulations. This combinatorics problem covers key concepts like binomial coefficients and is suitable for Grades 10-12.

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