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. **Write the given equation using formulas:** $$P(n, 2) = C(n+1, 3)$$ $$\frac{n!}{(n-2)!} = \frac{(n+1)!}{3!((n+1)-3)!}$$ 4. **Simplify both sides:** Left side: $$\frac{n!}{(n-2)!} = n \times (n-1)$$ Right side: $$\frac{(n+1)!}{3! (n-2)!} = \frac{(n+1) \times n \times (n-1) \times (n-2)!}{6 \times (n-2)!} = \frac{(n+1) \times n \times (n-1)}{6}$$ 5. **Set the simplified expressions equal:** $$n (n-1) = \frac{(n+1) n (n-1)}{6}$$ 6. **Cancel common factors $n (n-1)$ on both sides:** $$\cancel{n} \times \cancel{(n-1)} = \frac{(n+1) \times \cancel{n} \times \cancel{(n-1)}}{6}$$ $$1 = \frac{n+1}{6}$$ 7. **Solve for $n$:** $$6 = n + 1$$ $$n = 5$$ **Final answer:** $n = 5$