1. **State the problem:** Find the number of combinations of nickels and dimes that total 80 cents using exactly 11 coins. 2. **Define variables:** Let $n$ be the number of nickels and $d$ be the number of dimes. 3. **Write the system of equations:** - Total coins: $$n + d = 11$$ - Total value in cents: $$5n + 10d = 80$$ 4. **Solve the first equation for $n$:** $$n = 11 - d$$ 5. **Substitute $n$ into the value equation:** $$5(11 - d) + 10d = 80$$ 6. **Simplify:** $$55 - 5d + 10d = 80$$ $$55 + 5d = 80$$ 7. **Isolate $d$:** $$5d = 80 - 55$$ $$5d = 25$$ 8. **Divide both sides by 5:** $$\cancel{5}d = \cancel{5}5$$ $$d = 5$$ 9. **Find $n$ using $n = 11 - d$:** $$n = 11 - 5 = 6$$ 10. **Check the solution:** - Number of coins: $6 + 5 = 11$ - Total value: $5 \times 6 + 10 \times 5 = 30 + 50 = 80$ cents **Final answer:** There is exactly 1 combination: 6 nickels and 5 dimes.