Question

A box contains 5 blue, 6 yellow and 4 red balls. The number of ways, of drawing 8 balls containing at least two balls of each colour, is :

• A. 4100
• B. 4140
• C. 4230
• D. 4290

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We must select 8 balls such that Blue $\ge 2$, Yellow $\ge 2$, and Red $\ge 2$. We list all possible distributions $(b, y, r)$ that sum to 8.
: Key Formula or Approach:
Use $\binom{n}{r}$ for selection from each color.
Total balls drawn $= b + y + r = 8$.
Step 2: Detailed Explanation:
Cases for $(b, y, r)$:
1) $(2, 2, 4): \binom{5}{2} \cdot \binom{6}{2} \cdot \binom{4}{4} = 10 \cdot 15 \cdot 1 = 150$.
2) $(2, 4, 2): \binom{5}{2} \cdot \binom{6}{4} \cdot \binom{4}{2} = 10 \cdot 15 \cdot 6 = 900$.
3) $(4, 2, 2): \binom{5}{4} \cdot \binom{6}{2} \cdot \binom{4}{2} = 5 \cdot 15 \cdot 6 = 450$.
4) $(2, 3, 3): \binom{5}{2} \cdot \binom{6}{3} \cdot \binom{4}{3} = 10 \cdot 20 \cdot 4 = 800$.
5) $(3, 2, 3): \binom{5}{3} \cdot \binom{6}{2} \cdot \binom{4}{3} = 10 \cdot 15 \cdot 4 = 600$.
6) $(3, 3, 2): \binom{5}{3} \cdot \binom{6}{3} \cdot \binom{4}{2} = 10 \cdot 20 \cdot 6 = 1200$.
Total ways $= 150 + 900 + 450 + 800 + 600 + 1200 = 4100$.
Step 3: Final Answer:
Total ways is 4100.