Question:

Step 1: Understand the constraint on the selection. We need to select 18 balls from 4 colours (red, white, blue, green) such that there are at least 2 balls of each colour. Let $r$, $w$, $b$, and $g$ denote the number of red, white, blue, and green balls respectively. We require: $$r, w, b, g \geq 2 \quad \text{and} \quad r + w + b + g = 18$$ Step 2: Transform the problem using a change of variables. To convert this into a standard stars and bars problem, we introduce new variables to account for the minimum requirement. Let: $$r' = r - 2, \quad w' = w - 2, \quad b' = b - 2, \quad g' = g - 2$$ Since $r, w, b, g \geq 2$, we have $r', w', b', g' \geq 0$ (all non-negative integers). Step 3: Derive the equation in terms of new variables. Substituting the new variables into the original equation: $$(r' + 2) + (w' + 2) + (b' + 2) + (g' + 2) = 18$$ Simplifying: $$r' + w' + b' + g' + 8 = 18$$ $$r' + w' + b' + g' = 10$$ Step 4: Apply the stars and bars formula. The number of non-negative integer solutions to the equation $r' + w' + b' + g' = 10$ is given by the stars and bars formula: $$\binom{n + k - 1}{k - 1}$$ where $n = 10$ (the sum) and $k = 4$ (the number of variables). Therefore: $$k = \binom{10 + 4 - 1}{4 - 1} = \binom{13}{3}$$ Step 5: Calculate the binomial coefficient. $$\binom{13}{3} = \frac{13!}{3! \cdot 10!} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = \frac{1716}{6} = 286$$ **Final Answer:** The number of ways to select 18 balls with at least 2 of each colour is $k = 286$. This matches **Option 2: 286**.