JEE Advanced 1986

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Animated Solution for Mathematics - Permutations and Combinations: A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw?

Enter Numerical Value:

Visualized Solution

Inventory of the Box

Total balls in the box: $2 White + 3 Black + 4 Red = 9$ balls
We need to draw exactly $3$ balls.
Constraint: At least $1$ black ball must be included in the draw.

Classifying into Black and Non-Black

To simplify, we group the balls into two categories:
Black Balls: $3$
Non-Black Balls (White + Red): $2 + 4 = 6$

Strategy 1: Case-by-Case Analysis

At least one black ball means we can have:
Case 1: Exactly $1$ Black ball and $2$ Non-Black balls.
Case 2: Exactly $2$ Black balls and $1$ Non-Black ball.
Case 3: Exactly $3$ Black balls and $0$ Non-Black balls.

Case 1: Exactly $1$ Black and $2$ Non-Black

Select $1$ Black ball from $3$ available: $^{3} C_{1}$
Select $2$ Non-Black balls from $6$ available: $^{6} C_{2}$
Number of ways: $^{3} C_{1} \times^{6} C_{2}$
Calculation: $3 \times \frac{6 \times 5}{2 \times 1} = 3 \times 15 = 45$

Case 2: Exactly $2$ Black and $1$ Non-Black

Select $2$ Black balls from $3$ available: $^{3} C_{2}$
Select $1$ Non-Black ball from $6$ available: $^{6} C_{1}$
Number of ways: $^{3} C_{2} \times^{6} C_{1}$
Calculation: $3 \times 6 = 18$

Case 3: Exactly $3$ Black and $0$ Non-Black

Select $3$ Black balls from $3$ available: $^{3} C_{3}$
Select $0$ Non-Black balls from $6$ available: $^{6} C_{0}$
Number of ways: $^{3} C_{3} \times^{6} C_{0}$
Calculation: $1 \times 1 = 1$

Summing up the Cases

Total ways = Case 1 + Case 2 + Case 3
Total ways $= 45 + 18 + 1 = 64$

Strategy 2: The Complement Method

Total possible ways to select $3$ balls from $9$ without any restriction: $^{9} C_{3}$
Unfavorable ways (No black ball selected, i.e., all $3$ from $6$ non-black): $^{6} C_{3}$
Formula: $At least one Black = Total - No Black$

Calculating Total and Unfavorable Ways

Total ways $=^{9} C_{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$
Unfavorable ways (all $3$ from $6$ non-black) $=^{6} C_{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$

Final Result Verification

Ways with at least one black ball $= 84 - 20 = 64$
Both methods yield the exact same result: 64

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The Sigma Insight: Combinations and Selection

Solution Diagram

Analyzing the Setup

To solve this problem, we first categorize the nine spheres based on the condition provided. We have

3

black balls and

6

non-black balls (consisting of

2

white and

4

red).

Our objective is to select

3

balls such that at least one of them is black. This binary partition of

3

Black and

6

Non-Black simplifies our combinatorial space significantly.

The Direct Approach (Case-by-Case)

We can calculate the number of valid outcomes by summing three mutually exclusive scenarios.

Case 1: Exactly one black ball. We choose

1

black ball from

3

and

2

non-black balls from

6

^{3} C_{1} \times^{6} C_{2} = 3 \times 15 = 45

Case 2: Exactly two black balls. We choose

2

black balls from

3

and

1

non-black ball from

6

^{3} C_{2} \times^{6} C_{1} = 3 \times 6 = 18

Case 3: Exactly three black balls. We choose

3

black balls from

3

and

0

non-black balls from

6

^{3} C_{3} \times^{6} C_{0} = 1 \times 1 = 1

Summing these cases, we get

45 + 18 + 1 = 64

The Elegant Shortcut (Complement Method)

Alternatively, we can use the complement method by subtracting the "forbidden" cases (where no black balls are chosen) from the total possible combinations.

The total number of ways to choose

3

balls from

9

is:

^{9} C_{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84

The number of ways to choose

3

balls such that none are black (choosing all

3

from the

6

non-black balls) is:

^{6} C_{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20

Subtracting the forbidden cases from the total, we find the number of valid ways:

84 - 20 = 64

Conclusion

Both the direct case-by-case analysis and the complement method yield the same result. The total number of ways to select at least one black ball is 64.

Common Traps & FAQs

Wait, why can't I just pick one black ball first and then pick any two from the remaining eight balls? That seems easier.

That is a very common trap! If you pick one black ball first and then any two from the remaining, you will overcount. For example, if you pick Black Ball A first, then Black Ball B and Red Ball C, that is the same set as picking Black Ball B first, then Black Ball A and Red Ball C. The 'order' of selection doesn't matter in combinations, so picking one first creates duplicates. Always use combinations (

^{n} C_{r}

) to avoid this!

Why is the complement method (Total - Unfavorable) considered 'better'?

It is not necessarily 'better' in terms of accuracy, but it is often 'better' in terms of efficiency and reducing calculation errors. In problems with many cases (e.g., 'at least 5 black balls'), the direct method becomes tedious. The complement method turns a long, multi-step problem into a simple subtraction. It is a powerful tool to keep in your JEE toolkit.

Does the order in which I draw the balls matter? The question says 'drawn', not 'arranged'.

Excellent observation! You are absolutely right. Because the question asks for the number of ways to 'draw' a group of balls, the order of selection does not matter. This is why we use Combinations (

^{n} C_{r}

) rather than Permutations (

^{n} P_{r}

). If the question asked for the number of ways to 'arrange' them in a line, then order would matter.

Animated Solution for Mathematics - Permutations and Combinations: A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw?

Enter Numerical Value:

Visualized Solution

Inventory of the Box

Classifying into Black and Non-Black

Strategy 1: Case-by-Case Analysis

Case 1: Exactly $1$ Black and $2$ Non-Black

Case 2: Exactly $2$ Black and $1$ Non-Black

Case 3: Exactly $3$ Black and $0$ Non-Black

Summing up the Cases

Strategy 2: The Complement Method

Calculating Total and Unfavorable Ways

Final Result Verification

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Direct Approach (Case-by-Case)

The Elegant Shortcut (Complement Method)

Conclusion

Similar Questions

Consider 4 boxes, where each box contains 3 red balls and 2 blue balls. Assume that all 20 balls are distinct. In how many different ways can 10 balls be chosen from these 4 boxes so that from each box at least one red ball and one blue ball are chosen?

The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is

Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is :

The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is

An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at the most three of them are red is ....

Five balls of different colours are to be placed in there boxes of different size. Each box can hold all five. In how many different ways can we place the balls so that no box remains empty ?

An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at most three of them are red is __________.

The number of ways of giving 20 distinct oranges to 3 children such that each child gets at least one orange is _____

The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is ______.

Animated Solution for Mathematics - Permutations and Combinations: A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw?

Enter Numerical Value:

Visualized Solution

Inventory of the Box

Classifying into Black and Non-Black

Strategy 1: Case-by-Case Analysis

Case 1: Exactly .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1 Black and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2 Non-Black

Case 2: Exactly .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2 Black and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1 Non-Black

Case 3: Exactly .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3 Black and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 0 Non-Black

Summing up the Cases

Strategy 2: The Complement Method

Calculating Total and Unfavorable Ways

Final Result Verification

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Direct Approach (Case-by-Case)

The Elegant Shortcut (Complement Method)

Conclusion

Similar Questions

Consider 4 boxes, where each box contains 3 red balls and 2 blue balls. Assume that all 20 balls are distinct. In how many different ways can 10 balls be chosen from these 4 boxes so that from each box at least one red ball and one blue ball are chosen?

The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is

Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is :

The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is

An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at the most three of them are red is ....

Five balls of different colours are to be placed in there boxes of different size. Each box can hold all five. In how many different ways can we place the balls so that no box remains empty ?

An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at most three of them are red is __________.

The number of ways of giving 20 distinct oranges to 3 children such that each child gets at least one orange is _____

The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is ______.

Case 1: Exactly $1$ Black and $2$ Non-Black

Case 2: Exactly $2$ Black and $1$ Non-Black

Case 3: Exactly $3$ Black and $0$ Non-Black