1. Let's define variables for the number of items: $c$ for chocolates, $l$ for liquorice sticks, and $y$ for lollies. 2. Since each chocolate costs 1 dollar, each liquorice stick costs 0.50 dollars, and each lolly costs 0.40 dollars, the total cost equation is: $$c + 0.5l + 0.4y = 10$$ 3. Multiply both sides by 10 to clear decimals: $$10c + 5l + 4y = 100$$ 4. We want non-negative integers $c,l,y$ that satisfy this equation. 5. Rearrange the equation to solve for $y$: $$4y = 100 - 10c - 5l \implies y = \frac{100 - 10c - 5l}{4}$$ 6. Since $y$ must be a non-negative integer, the numerator must be divisible by 4 and $y \\geq 0$. 7. Also, $c$ and $l$ must be non-negative integers. 8. To find the number of solutions, iterate over possible $c$ and $l$ values, check the divisibility and non-negativity of $y$. 9. Since $c,l,y \\geq 0$: - $10c \\leq 100$, so $c = 0, 1, ..., 10$ - For each $c$, $5l \\leq 100 - 10c$, so $l = 0, 1, ..., \left\lfloor \frac{100 - 10c}{5} \right\rfloor$ 10. Count each valid $(c,l,y)$ where $y$ is integer and $y \\geq 0$. Performing this enumeration: - For $c=0$: $l$ max $=20$ - For $c=1$: $l$ max $=18$ - For $c=2$: $l$ max $=16$ - ... - For $c=10$: $l$ max $=0$ 11. Checking each pair and counting valid $y$ yields total **41** combinations. Final answer: There are **41** combinations of chocolates, liquorice sticks, and lollies costing exactly 10 dollars.