1. **State the problem:** Evaluate the expression $$\frac{17!}{3!(17-3)!}$$ and determine if the result is an integer or not. 2. **Recall the formula:** This expression matches the formula for combinations, which is $$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$ where $n=17$ and $r=3$. 3. **Apply the formula:** Substitute values: $$\binom{17}{3} = \frac{17!}{3!(17-3)!} = \frac{17!}{3!14!}$$ 4. **Simplify the factorial expression:** Expand $17!$ as $17 \times 16 \times 15 \times 14!$ to cancel $14!$ in numerator and denominator: $$\frac{17 \times 16 \times 15 \times 14!}{3! \times 14!} = \frac{17 \times 16 \times 15 \times \cancel{14!}}{3! \times \cancel{14!}}$$ 5. **Calculate $3!$:** $$3! = 3 \times 2 \times 1 = 6$$ 6. **Rewrite the expression:** $$\frac{17 \times 16 \times 15}{6}$$ 7. **Simplify the fraction:** Divide numerator and denominator by 3: $$\frac{17 \times \cancel{16} \times 15}{\cancel{6}} = \frac{17 \times 16 \times 15}{6}$$ Actually, better to cancel stepwise: $$\frac{17 \times 16 \times 15}{6} = 17 \times 16 \times \frac{15}{6}$$ Simplify $\frac{15}{6}$: $$\frac{15}{6} = \frac{\cancel{15}^{5}}{\cancel{6}^{2}} = \frac{5}{2}$$ 8. **Calculate the product:** $$17 \times 16 \times \frac{5}{2} = 17 \times 16 \times 2.5$$ Calculate $16 \times 2.5$: $$16 \times 2.5 = 40$$ Then multiply by 17: $$17 \times 40 = 680$$ 9. **Conclusion:** The value is an integer and equals 680. **Final answer:** $$\frac{17!}{3!(17-3)!} = 680$$