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.