Differentiate between permutation and combination. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls can be drawn from the box such that at least one ball…

Difference Between Permutation and Combination

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Solution: Drawing 3 Balls with At Least One Black Ball

Given:

• 2 white balls (W)
• 3 black balls (B)
• 4 red balls (R)
• Total balls = 2 + 3 + 4 = 9

Objective: Find the number of ways to draw 3 balls such that at least one is black.

Method 1: Direct Counting (Using Combinations)

We calculate the number of ways to draw 3 balls including at least one black ball by considering cases where:

1. Exactly 1 black ball is drawn.
2. Exactly 2 black balls are drawn.
3. All 3 black balls are drawn.

Case 1: Exactly 1 Black Ball

• Choose 1 black ball from 3: $C(3,1)$
• Choose 2 non-black balls from remaining 6 (2W + 4R): $C(6,2)$
• Total ways: $C(3,1)\times C(6,2)=3\times 15=45$

Case 2: Exactly 2 Black Balls

• Choose 2 black balls from 3: $C(3,2)$
• Choose 1 non-black ball from 6: $C(6,1)$
• Total ways: $C(3,2)\times C(6,1)=3\times 6=18$

Case 3: All 3 Black Balls

• Choose all 3 black balls: $C(3,3)=1$
• No non-black balls are chosen.

Total Favorable Outcomes: $$45+18+1=64$$

Method 2: Complementary Counting (Easier)

Calculate the total number of ways to draw 3 balls without any restrictions and subtract the cases where no black balls are drawn.

1. Total ways to draw 3 balls from 9: $$C(9,3)=\frac{9!}{3!6!}=84$$

2. Ways to draw 3 balls with no black balls (only from 6 non-black balls): $$C(6,3)=20$$

3. Favorable outcomes (at least one black ball): $$84-20=64$$

Final Answer: The number of ways to draw 3 balls with at least one black ball is 64.