JEE Main 2020 - 5 Sep (Evening)

LEVELJEE Main

Animated Solution for Mathematics - Permutations and Combinations: There are 3 sections in a question paper and each section contains 5 questions. A candidate has to answer a total of 5 questions, choosing at least one question from each section. Then the number of ways, in which the candidate can choose the questions, is :

Q: Why do we multiply by $3$ in the permutation step?

Think of it this way: the distribution pattern $(1, 1, 3)$ means one section contributes $3$ questions, while the other two contribute $1$ each. There are $3$ different sections that could be the one contributing $3$ questions. Therefore, there are $3$ distinct ways to assign the $(1, 1, 3)$ pattern to the sections (S1, S2, S3). We multiply by $3$ to account for all these possible assignments.

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the problem as a distribution task. You have

5

total questions to select from

3

distinct sections, with the constraint that each section must contribute at least

1

question. This is equivalent to partitioning the integer

5

into

3

positive parts, where the order of the parts matters because the sections are distinct.

Hint 2

Hint 2 (The Arsenal): Use the concept of combinations

^{n} C_{r}

to select questions within each section. Remember that the distribution of the

5

questions across the

3

sections can happen in different patterns (e.g.,

1, 1, 3

1, 2, 2

). For each pattern, calculate the number of ways to choose the questions, and then multiply by the number of ways to permute that pattern across the sections.

Hint 3

Hint 3 (The Execution): Start by identifying the two possible integer partitions of

5

into

3

positive parts:

(1, 1, 3)

and

(1, 2, 2)

. For the

(1, 1, 3)

case, calculate

^{5} C_{1} \times^{5} C_{1} \times^{5} C_{3}

and multiply by the number of permutations of the set

(1, 1, 3)

. Repeat this process for the

(1, 2, 2)

case using

^{5} C_{1} \times^{5} C_{2} \times^{5} C_{2}

and its permutations, then sum the results.

Visualized Solution

The Question Paper Structure

Total sections: $3$ (S1, S2, S3)
Questions per section: $5$
Total available questions: $15$

The Selection Constraint

Total questions to answer: $5$
Constraint: At least one question from each section
Let $x_{i}$ be the number of questions chosen from section $i$
$x_{1} + x_{2} + x_{3} = 5$ , where $x_{i} \geq 1$

Integer Partitions of $5$

We need to partition $5$ into $3$ positive integers.
Possible sets of $(x_{1}, x_{2}, x_{3})$ :
Case 1: $(1, 1, 3)$
Case 2: $(1, 2, 2)$

Case 1: The $(1, 1, 3)$ Distribution

Assume we select $1$ from S1, $1$ from S2, and $3$ from S3.
Ways to choose from S1: $^{5} C_{1}$
Ways to choose from S2: $^{5} C_{1}$
Ways to choose from S3: $^{5} C_{3}$

Evaluating $(1, 1, 3)$ Assignment

Number of ways = $^{5} C_{1} \times^{5} C_{1} \times^{5} C_{3}$
$^{5} C_{1} = 5$
$^{5} C_{3} = \frac{5 \times 4}{2} = 10$
Ways = $5 \times 5 \times 10 = 250$

Permuting the $(1, 1, 3)$ Pattern

The pattern $(1, 1, 3)$ can be assigned to (S1, S2, S3) in multiple ways.
Number of permutations = $\frac{3 !}{2 !} = 3$
The ' $3$ ' questions could be from S1, S2, or S3.

Total Ways for Case 1

Total ways for Case 1 = (Ways for one arrangement) $\times$ (Number of arrangements)
Total ways = $250 \times 3 = 750$

Case 2: The $(1, 2, 2)$ Distribution

Now consider selecting $1$ from S1, $2$ from S2, and $2$ from S3.
Ways to choose from S1: $^{5} C_{1}$
Ways to choose from S2: $^{5} C_{2}$
Ways to choose from S3: $^{5} C_{2}$

Evaluating $(1, 2, 2)$ Assignment

Number of ways = $^{5} C_{1} \times^{5} C_{2} \times^{5} C_{2}$
$^{5} C_{1} = 5$
$^{5} C_{2} = 10$
Ways = $5 \times 10 \times 10 = 500$

Permuting the $(1, 2, 2)$ Pattern

The pattern $(1, 2, 2)$ can also be shuffled among the $3$ sections.
Number of permutations = $\frac{3 !}{2 !} = 3$
The ' $1$ ' question could be from S1, S2, or S3.

Total Ways for Case 2

Total ways for Case 2 = (Ways for one arrangement) $\times$ (Number of arrangements)
Total ways = $500 \times 3 = 1500$

Total Number of Ways

Total ways = (Total ways for Case 1) + (Total ways for Case 2)
Total ways = $750 + 1500$
Total ways = $2250$

00:00 / 00:00

The Sigma Insight: Combinations and Selection

Solution Diagram

Analyzing the Setup

We have

3

distinct sections,

S_{1}, S_{2},

and

S_{3}

, each containing

5

questions. We are tasked with selecting a total of

5

questions such that at least one question is chosen from each section.

Let

x_{1}, x_{2},

and

x_{3}

represent the number of questions selected from sections

S_{1}, S_{2},

and

S_{3}

respectively. We must satisfy the equation:

x_{1} + x_{2} + x_{3} = 5, where x_{i} \geq 1

This is a partition problem where we distribute the integer

5

into

3

positive parts. The possible integer partitions are

(3, 1, 1)

and

(2, 2, 1)

Case 1

The

(3, 1, 1)

Distribution

In this scenario, one section contributes

3

questions, while the other two sections contribute

1

question each.

First, we calculate the number of ways for a fixed assignment, such as

x_{1} = 3, x_{2} = 1, x_{3} = 1

(3 5) \times (1 5) \times (1 5) = 10 \times 5 \times 5 = 250

Since the section providing

3

questions can be any of the

3

sections, we multiply by the number of permutations, which is

\frac{3 !}{2 !} = 3

The total number of ways for this case is:

250 \times 3 = 750

Case 2

The

(2, 2, 1)

Distribution

In this scenario, one section contributes

1

question, while the other two sections contribute

2

questions each.

For a fixed assignment, such as

x_{1} = 1, x_{2} = 2, x_{3} = 2

, the number of ways is:

(1 5) \times (2 5) \times (2 5) = 5 \times 10 \times 10 = 500

The section providing the single question can be chosen in

3

different ways (the permutations of the set

{1, 2, 2}

The total number of ways for this case is:

500 \times 3 = 1500

Final Calculation

To find the total number of valid combinations, we sum the results from both mutually exclusive cases:

Total Ways = 750 + 1500 = 2250

The total number of ways to choose the questions under the given constraints is 2250.

Common Traps & FAQs

Wait, why can't we just use the stars and bars method here?

That is a fantastic question! The stars and bars method is designed for distributing identical items into distinct bins. However, in this problem, the questions themselves are distinct entities within each section. Choosing question A from Section 1 is different from choosing question B from Section 1. Because the items are distinct, we must use combinations (

^{n} C_{r}

) to account for the specific questions chosen, rather than just the count of questions.

Why do we multiply by

3

in the permutation step?

Think of it this way: the distribution pattern

(1, 1, 3)

means one section contributes

3

questions, while the other two contribute

1

each. There are

3

different sections that could be the one contributing

3

questions. Therefore, there are

3

distinct ways to assign the

(1, 1, 3)

pattern to the sections (S1, S2, S3). We multiply by

3

to account for all these possible assignments.

Is there a way to solve this using the principle of inclusion-exclusion?

Yes, absolutely! You could calculate the total ways to choose

5

questions from

15

(

^{15} C_{5}

) and then subtract the cases where one or more sections are empty. However, that approach is often more prone to calculation errors due to the complexity of the overlapping sets. The case-based approach we used is much more intuitive and less likely to lead to double-counting.

Select Answer:

Visualized Solution

The Question Paper Structure

The Selection Constraint

Integer Partitions of $5$

Case 1: The $(1, 1, 3)$ Distribution

Evaluating $(1, 1, 3)$ Assignment

Permuting the $(1, 1, 3)$ Pattern

Total Ways for Case 1

Case 2: The $(1, 2, 2)$ Distribution

Evaluating $(1, 2, 2)$ Assignment

Permuting the $(1, 2, 2)$ Pattern

Total Ways for Case 2

Total Number of Ways

The Sigma Insight: Combinations and Selection

Analyzing the Setup

Case 1

Case 2

Final Calculation

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Select Answer:

Visualized Solution

The Question Paper Structure

The Selection Constraint

Integer Partitions of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5

Case 1: The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1,1,3) Distribution

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1,1,3) Assignment

Permuting the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1,1,3) Pattern

Total Ways for Case 1

Case 2: The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1,2,2) Distribution

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1,2,2) Assignment

Permuting the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1,2,2) Pattern

Total Ways for Case 2

Total Number of Ways

The Sigma Insight: Combinations and Selection

Analyzing the Setup

Case 1

Case 2

Final Calculation

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A boy needs to select five courses from 12 available courses, out of which 5 courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is ____.

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Integer Partitions of $5$

Case 1: The $(1, 1, 3)$ Distribution

Evaluating $(1, 1, 3)$ Assignment

Permuting the $(1, 1, 3)$ Pattern

Case 2: The $(1, 2, 2)$ Distribution

Evaluating $(1, 2, 2)$ Assignment

Permuting the $(1, 2, 2)$ Pattern