Let’s solve this step by step in a simple and fun way! 🎉
Step 1: Identify the given balls ⚪⚫🔴
The box contains:

• 2 White balls ⚪⚪
• 3 Black balls ⚫⚫⚫
• 4 Red balls 🔴🔴🔴🔴

We need to draw 3 balls with the condition that at least one black ball must be included. ✅

Step 2: Find total ways to draw any 3 balls
There are a total of 2 + 3 + 4 = 9 balls.
The total ways to select any 3 balls (without restriction) is:

(93)=9!3!(9−3)!=9×8×73×2×1=84\binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84

Step 3: Find the ways to select 3 balls without any black ball ❌⚫
If we ignore black balls, we only have White (2) + Red (4) = 6 balls.
Ways to pick 3 balls from these 6:

(63)=6!3!(6−3)!=6×5×43×2×1=20\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20

Step 4: Apply the condition (At least 1 black ball) ✅⚫
Since we need at least one black ball, we subtract the cases where there are no black balls:

Total ways−Ways with no black ball=84−20=64\text{Total ways} – \text{Ways with no black ball} = 84 – 20 = 64

🎯 Final Answer:
The number of ways to draw 3 balls while ensuring at least one black ball is included is 64! 🎊😊