Question

A bag contains 4 red balls, 6 yellow balls and 5 blue balls. In how many ways we can select 8 balls such that we get at least two balls of each colour?

Answer 1

Correct Answer: 4100

Step 1: Understanding the Concept:
We need to select a total of 8 balls from a set of 15 (4R, 6Y, 5B) with the constraint that each color must have at least 2 balls. This involves identifying all possible combinations of counts (r, y, b) such that r + y + b = 8 and each is at least 2.

Step 2: Key Formula or Approach:
For each valid combination (r, y, b), the number of ways is calculated using combinations: C(4, r) × C(6, y) × C(5, b).

Step 3: Detailed Explanation:
Possible combinations (r, y, b):

(2, 2, 4): C(4,2) × C(6,2) × C(5,4) = 6 × 15 × 5 = 450
(2, 3, 3): C(4,2) × C(6,3) × C(5,3) = 6 × 20 × 10 = 1200
(2, 4, 2): C(4,2) × C(6,4) × C(5,2) = 6 × 15 × 10 = 900
(3, 2, 3): C(4,3) × C(6,2) × C(5,3) = 4 × 15 × 10 = 600
(3, 3, 2): C(4,3) × C(6,3) × C(5,2) = 4 × 20 × 10 = 800
(4, 2, 2): C(4,4) × C(6,2) × C(5,2) = 1 × 15 × 10 = 150

Total ways = 450 + 1200 + 900 + 600 + 800 + 150 = 4100.

Step 4: Final Answer:
The total number of ways to select the balls is 4100.