1. Let's state the problem: We want to find the number of combinations of chocolates, liquorice sticks, and lollies that total 10 dollars. 2. Define variables: Let $x$ be the number of chocolates, $y$ be the number of liquorice sticks, and $z$ be the number of lollies. 3. Write the cost equation: Each chocolate costs 1 dollar, so cost from chocolates is $1 \cdot x = x$. Each liquorice stick costs 50 cents, or 0.5 dollars, so cost from liquorice sticks is $0.5 \cdot y$. Each lolly costs 40 cents, or 0.4 dollars, so cost from lollies is $0.4 \cdot z$. 4. Equation for total cost: $$x + 0.5y + 0.4z = 10$$ 5. Multiply through by 10 to clear decimals: $$10x + 5y + 4z = 100$$ 6. We want nonnegative integer solutions $(x,y,z \geq 0)$ satisfying: $$10x + 5y + 4z = 100$$ 7. For fixed $x$, rearrange to find $y$ and $z$: $$5y + 4z = 100 - 10x$$ 8. For $x=0$ to $10$ (since $10x \le 100$): - For each $x$, set $R = 100 - 10x$. - For $y=0$ to $\lfloor R/5 \rfloor$, check if $4z = R - 5y$ is divisible by 4. - If yes, $z = \frac{R - 5y}{4}$ is a nonnegative integer. 9. Count all such $(x,y,z)$. 10. Let's compute: - For $x=0$, $R=100$: $y$ from 0 to 20: Solutions where $R-5y$ divisible by 4: Values of $y$ that satisfy $100 - 5y \equiv 0 \pmod{4}$. Check $100-5y \equiv 0 \pmod{4}$: Since $100 \equiv 0 \pmod{4}$, the condition becomes $-5y \equiv 0 \pmod{4}$ or equivalently $5y \equiv 0 \pmod{4}$. Since $5 \equiv 1 \pmod{4}$, this reduces to $y \equiv 0 \pmod{4}$. Thus for $x=0$, $y = 0,4,8,12,16,20$. Number of $y$ values: 6. For each $y$, compute $z = (100 - 5y)/4$ will be nonnegative integers. Similarly, for $x=1$ to $10$, repeat the same process: Generalizing: $R = 100 - 10x$ Want $y \equiv 0 \pmod{4}$ and $0 \leq y \leq \left\lfloor R/5 \right\rfloor$. Count $y$ satisfying these constraints, then $z$ is uniquely determined. Calculate counts: x=0, R=100, max y = 20, y values: 0,4,8,12,16,20 (6 values) x=1, R=90, max y=18: y=0,4,8,12,16 (5 values) x=2, R=80, max y=16: y=0,4,8,12,16 (5 values) x=3, R=70, max y=14: y=0,4,8,12 (4 values) x=4, R=60, max y=12: y=0,4,8,12 (4 values) x=5, R=50, max y=10: y=0,4,8 (3 values) x=6, R=40, max y=8: y=0,4,8 (3 values) x=7, R=30, max y=6: y=0,4 (2 values) x=8, R=20, max y=4: y=0,4 (2 values) x=9, R=10, max y=2: y=0 (1 value) x=10,R=0, max y=0: y=0 (1 value) Sum all counts: 6+5+5+4+4+3+3+2+2+1+1 = 36 11. Final answer: There are 36 combinations of chocolates, liquorice sticks, and lollies costing exactly 10 dollars.