JEE Main 2021 (25 July Shift 1)

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Animated Solution for Mathematics - Permutations and Combinations: There are 5 students in class 10, 6 students in class 11 and 8 students in class 12. If the number of ways, in which 10 students can be selected from them so as to include at least 2 students from each class and at most 5 students from the total 11 students of class 10 and 11 is 100k, then k is equal to

Q: Why is the sum of $n_{10} + n_{11}$ only 4 or 5? Why not 6?

Great observation! The constraint says $n_{10} + n_{11} \le 5$. If the sum were 6, it would violate the 'at most 5' rule. Furthermore, since $n_{10} \ge 2$ and $n_{11} \ge 2$, the minimum sum is $2+2=4$. Thus, the sum is strictly trapped between 4 and 5.

Enter Numerical Value:

Progressive Hints

Hint 1

The Spark: Do not rush into the calculation. First, visualize the constraints. You have a total of 10 students to select, but you are restricted by a 'floor' (at least 2 from each class) and a 'ceiling' (at most 5 from the combined pool of Class 10 and 11). Focus on the sum

n_{10} + n_{11}

. What is the minimum and maximum value this sum can take given the constraints?

Hint 2

The Arsenal: Once you identify that the sum

n_{10} + n_{11}

can only be 4 or 5, you have effectively partitioned the problem into two mutually exclusive cases. Use the fundamental counting principle: for each case, calculate the number of ways to choose students from each class independently and multiply them together.

Hint 3

The Execution: For Case 1, where

n_{10} + n_{11} = 4

, you must have

n_{10} = 2

and

n_{11} = 2

. For Case 2, where

n_{10} + n_{11} = 5

, you have two sub-scenarios:

(n_{10} = 2, n_{11} = 3)

and

(n_{10} = 3, n_{11} = 2)

. Set up your combinations as

n_{10} C_{r} \times n_{11} C_{r} \times n_{12} C_{r}

for each scenario and sum them up.

Visualized Solution

Defining the Variables

Let $n_{10}$ , $n_{11}$ , $n_{12}$ be the number of students selected from Class 10, 11, and 12.
Total students available: $5$ , $6$ , and $8$ respectively.
Total students to be selected: $n_{10} + n_{11} + n_{12} = 10$ .

Establishing the Constraints

Constraint 1: At least $2$ students from each class.
$n_{10} \geq 2, n_{11} \geq 2, n_{12} \geq 2$
Constraint 2: At most $5$ students from Class 10 and 11 combined.
$n_{10} + n_{11} \leq 5$

Finding the Range for $n_{10} + n_{11}$

From Constraint 1: $n_{10} \geq 2$ and $n_{11} \geq 2 ⟹ n_{10} + n_{11} \geq 4$ .
From Constraint 2: $n_{10} + n_{11} \leq 5$ .
Therefore, the possible values for $n_{10} + n_{11}$ are exactly $4$ or $5$ .

Case 1: $n_{10} + n_{11} = 4$

If $n_{10} + n_{11} = 4$ , then $n_{12} = 10 - 4 = 6$ .
Since $n_{10} \geq 2$ and $n_{11} \geq 2$ , the only possible selection is:
$n_{10} = 2$ and $n_{11} = 2$ .
Selection tuple $(n_{10}, n_{11}, n_{12}) = (2, 2, 6)$ .

Calculating Ways for Case 1

Number of ways = $^{5} C_{2} \times^{6} C_{2} \times^{8} C_{6}$
$^{5} C_{2} = 10$ , $^{6} C_{2} = 15$ , $^{8} C_{6} = \frac{8 \times 7}{2 \times 1} = 28$
Total ways for Case 1 = $10 \times 15 \times 28 = 4200$ .

Case 2: $n_{10} + n_{11} = 5$

If $n_{10} + n_{11} = 5$ , then $n_{12} = 10 - 5 = 5$ .
Possible subcases for $(n_{10}, n_{11})$ :
Subcase A: $(2, 3)$
Subcase B: $(3, 2)$

Calculating Ways for Case 2

Ways for Subcase A: $^{5} C_{2} \times^{6} C_{3} \times^{8} C_{5} = 10 \times 20 \times 56 = 11200$
Ways for Subcase B: $^{5} C_{3} \times^{6} C_{2} \times^{8} C_{5} = 10 \times 15 \times 56 = 8400$
Total ways for Case 2 = $11200 + 8400 = 19600$ .

Total Ways and Solving for $k$

Total ways = Case 1 + Case 2
Total ways = $4200 + 19600 = 23800$
We are given that total ways = $100 k$ .
$100 k = 23800 ⟹ k = 238$ .

Final Conclusion

Key Takeaway:
Identify constraints clearly: $n_{i} \geq 2$ and $\sum n_{10, 11} \leq 5$ .
Use the constraints to limit the number of cases.
Calculate each case using $^{n} C_{r}$ and sum them up.
Final Answer: $k = 238$

00:00 / 00:00

The Sigma Insight: Combinations and Selection

Solution Diagram

Analyzing the Setup

We are tasked with selecting 10 students from three classes: Class 10 (5 students), Class 11 (6 students), and Class 12 (8 students). Let

n_{10}

n_{11}

, and

n_{12}

represent the number of students selected from each respective class.

The fundamental constraint is the total number of students selected:

n_{10} + n_{11} + n_{12} = 10

We are given the "floor" constraints, requiring at least 2 students from each class:

n_{10} \geq 2, n_{11} \geq 2, n_{12} \geq 2

The Master Constraint

We are also given a "ceiling" constraint regarding the combined selection from Class 10 and Class 11:

n_{10} + n_{11} \leq 5

Combining the floor constraints, we know

n_{10} + n_{11} \geq 2 + 2 = 4

. Since the sum must be at least 4 and no more than 5, the only possible values for the sum

n_{10} + n_{11}

are 4 or 5.

Case 1: $n_{10} + n_{11} = 4$

n_{10} + n_{11} = 4

, then

n_{12} = 10 - 4 = 6

. Given the constraints

n_{10} \geq 2

and

n_{11} \geq 2

, the only valid distribution is

n_{10} = 2

and

n_{11} = 2

The number of ways to choose these students is:

^{5} C_{2} \times^{6} C_{2} \times^{8} C_{6} = 10 \times 15 \times 28 = 4200

Case 2: $n_{10} + n_{11} = 5$

n_{10} + n_{11} = 5

, then

n_{12} = 10 - 5 = 5

. To satisfy

n_{10} \geq 2

and

n_{11} \geq 2

, we have two subcases:

Subcase A:

n_{10} = 2, n_{11} = 3

^{5} C_{2} \times^{6} C_{3} \times^{8} C_{5} = 10 \times 20 \times 56 = 11200

Subcase B:

n_{10} = 3, n_{11} = 2

^{5} C_{3} \times^{6} C_{2} \times^{8} C_{5} = 10 \times 15 \times 56 = 8400

The total for Case 2 is

11200 + 8400 = 19600

Final Calculation

Summing the results from both cases, we find the total number of ways:

4200 + 19600 = 23800

Given the problem states the total is

100 k

, we set

100 k = 23800

. Therefore, the value of $k = 238$ .

Common Traps & FAQs

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

Stars and bars is a powerful tool for distributing identical items into distinct bins, but it fails here because our students are distinct individuals, and we have specific constraints on the number of students from each class. Combinations (

^{n} C_{r}

) are the correct tool when selecting distinct items where order doesn't matter.

Why is the sum of

n_{10} + n_{11}

only 4 or 5? Why not 6?

Great observation! The constraint says

n_{10} + n_{11} \leq 5

. If the sum were 6, it would violate the 'at most 5' rule. Furthermore, since

n_{10} \geq 2

and

n_{11} \geq 2

, the minimum sum is

2 + 2 = 4

. Thus, the sum is strictly trapped between 4 and 5.

Do I need to worry about the order of selection?

No, you don't! The problem asks for the number of ways to 'select' students. In selection problems, the order in which you pick the students does not matter, which is exactly why we use combinations (

^{n} C_{r}

) rather than permutations (

^{n} P_{r}

Enter Numerical Value:

Visualized Solution

Defining the Variables

Establishing the Constraints

Finding the Range for $n_{10} + n_{11}$

Case 1: $n_{10} + n_{11} = 4$

Calculating Ways for Case 1

Case 2: $n_{10} + n_{11} = 5$

Calculating Ways for Case 2

Total Ways and Solving for $k$

Final Conclusion

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Master Constraint

Case 1: $n_{10} + n_{11} = 4$

Case 2: $n_{10} + n_{11} = 5$

Final Calculation

Similar Questions

There are ten boys $B_{1}, B_{2}, \dots, B_{10}$ and five girls $G_{1}, G_{2}, \dots, G_{5}$ in a class. Then the number of ways of forming a group consisting of three boys and three girls, if both $B_{1}$ and $B_{2}$ together should not be the members of a group, is ______.

A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is 1750, then n is equal to :

Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is:

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

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 ____.

A class contains $b$ boys and $g$ girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168, then $b + 3 g$ is equal to ______.

A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is

At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be selected, if a voter votes for at least one candidate, then the number of ways in which he can vote is

There are 15 players in a cricket team, out of which 6 are bowlers, 7 are batsmen and 2 are wicketkeepers. The number of ways, a team of 11 players be selected from them so as to include at least 4 bowlers, 5 batsmen and 1 wicketkeeper, is ___

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

Enter Numerical Value:

Visualized Solution

Defining the Variables

Establishing the Constraints

Finding the Range for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n10​+n11​

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n10​+n11​=4

Calculating Ways for Case 1

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n10​+n11​=5

Calculating Ways for Case 2

Total Ways and Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Final Conclusion

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Master Constraint

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n10​+n11​=4

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n10​+n11​=5

Final Calculation

Similar Questions

A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is 1750, then n is equal to :

Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is:

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

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 ____.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A class contains b boys and g girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168, then b+3g is equal to ______.

A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is

At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be selected, if a voter votes for at least one candidate, then the number of ways in which he can vote is

There are 15 players in a cricket team, out of which 6 are bowlers, 7 are batsmen and 2 are wicketkeepers. The number of ways, a team of 11 players be selected from them so as to include at least 4 bowlers, 5 batsmen and 1 wicketkeeper, is ___

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

Finding the Range for $n_{10} + n_{11}$

Case 1: $n_{10} + n_{11} = 4$

Case 2: $n_{10} + n_{11} = 5$

Total Ways and Solving for $k$

Case 1: $n_{10} + n_{11} = 4$

Case 2: $n_{10} + n_{11} = 5$

A class contains $b$ boys and $g$ girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168, then $b + 3 g$ is equal to ______.