Question

A bag contains 5 red balls, 6 blue balls and 4 black balls (balls of same colour are considered to be distinct). The number of ways in which 8 balls can be selected if at least two balls of each colour is there, is:

• A. 4250
• B. 3650
• C. 3200
• D. 4100

Hint:

Always check if items are "identical" or "distinct." If they were identical, we would use the stars and bars method with constraints; since they are distinct, we use $\binom{n}{r}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Since the balls of the same color are considered distinct, we use the combination formula $\binom{n}{r}$ for each color. We need to select 8 balls $(r, b, k)$ such that $r+b+k=8$ with the constraints $r \ge 2$, $b \ge 2$, and $k \ge 2$.

Step 2: Key Formula or Approach:
Identify the possible distributions of 8 balls among the three colors: 1. $(2, 2, 4)$ 2. $(2, 4, 2)$ 3. $(4, 2, 2)$ 4. $(2, 3, 3)$ 5. $(3, 2, 3)$ 6. $(3, 3, 2)$

Step 3: Detailed Explanation:
1. $(2, 2, 4): \binom{5}{2} \binom{6}{2} \binom{4}{4} = 10 \times 15 \times 1 = 150$
2. $(2, 4, 2): \binom{5}{2} \binom{6}{4} \binom{4}{2} = 10 \times 15 \times 6 = 900$
3. $(4, 2, 2): \binom{5}{4} \binom{6}{2} \binom{4}{2} = 5 \times 15 \times 6 = 450$
4. $(2, 3, 3): \binom{5}{2} \binom{6}{3} \binom{4}{3} = 10 \times 20 \times 4 = 800$
5. $(3, 2, 3): \binom{5}{3} \binom{6}{2} \binom{4}{3} = 10 \times 15 \times 4 = 600$
6. $(3, 3, 2): \binom{5}{3} \binom{6}{3} \binom{4}{2} = 10 \times 20 \times 6 = 1200$
Total ways $= 150 + 900 + 450 + 800 + 600 + 1200 = 4100$.

Step 4: Final Answer:
The number of ways is 4100.