1. **Stating the problem:** Given the equation $P(n, 2) = C(n + 1, 3)$, find the value of $n$ that satisfies this. 2. **Recall formulas:** - Permutation formula: $P(n, r) = \frac{n!}{(n-r)!}$ - Combination formula: $C(n, r) = \frac{n!}{r!(n-r)!}$ 3. **Apply the formulas:** $$P(n, 2) = \frac{n!}{(n-2)!} = n \times (n-1)$$ $$C(n+1, 3) = \frac{(n+1)!}{3! (n-2)!} = \frac{(n+1) n (n-1)}{6}$$ 4. **Set the equation:** $$n (n-1) = \frac{(n+1) n (n-1)}{6}$$ 5. **Simplify by canceling common factors:** Since $n (n-1)$ appears on both sides and assuming $n \neq 0$ and $n \neq 1$, we can cancel: $$\cancel{n} \times \cancel{(n-1)} = \frac{(n+1) \cancel{n} \times \cancel{(n-1)}}{6}$$ which simplifies to: $$1 = \frac{n+1}{6}$$ 6. **Solve for $n$:** Multiply both sides by 6: $$6 = n + 1$$ Subtract 1: $$n = 5$$ 7. **Answer:** The value of $n$ that satisfies the equation is $\boxed{5}$.