JEE Advanced 2016

LEVELJEE Main

Animated Solution for Mathematics - Permutations and Combinations: 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

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Progressive Hints

Hint 1

The Spark: Visualize the team as a collection of individuals. The constraint 'at most one boy' is a filter that restricts the possible compositions of the team. You must consider the scenarios where the number of boys is

0

and where it is

1

Hint 2

The Arsenal: Use the combination formula

^{n} C_{r} = \frac{n !}{r ! ( n - r )!}

to select members. Remember the Fundamental Principle of Counting: when you have mutually exclusive cases, you add the results; when you have sequential steps (like selecting the team and then the captain), you multiply.

Hint 3

The Execution: Calculate the number of ways to form the team for Case 1 (

0

boys,

4

girls) and Case 2 (

1

boy,

3

girls). Sum these to get the total team combinations. Finally, multiply this total by the number of ways to choose a captain from the

4

selected members.

Visualized Solution

The Starting Pool

Total members in the club: $6$ Girls and $4$ Boys.
Team size to be selected: $4$ members.
Constraint: The team must include at most one boy.

Defining 'At Most One Boy'

'At most one boy' implies two mutually exclusive cases:
Case 1: $0$ boys and $4$ girls.
Case 2: $1$ boy and $3$ girls.

Case 1: Selecting $4$ Girls

Number of ways to select $4$ girls from $6$ : $^{6} C_{4}$
Using the property $^{n} C_{r} =^{n} C_{n - r}$ , $^{6} C_{4} =^{6} C_{2}$ .

Calculating $^{6} C_{4}$

$^{6} C_{4} = \frac{6 \times 5}{2 \times 1}$
$^{6} C_{4} = 15$ ways.

Case 2: $1$ Boy and $3$ Girls

Number of ways to select $1$ boy from $4$ : $^{4} C_{1}$
Number of ways to select $3$ girls from $6$ : $^{6} C_{3}$
Total ways for Case 2: $^{4} C_{1} \times^{6} C_{3}$

Calculating $^{4} C_{1} \times^{6} C_{3}$

$^{4} C_{1} = 4$
$^{6} C_{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$
Total ways for Case 2: $4 \times 20 = 80$ ways.

Total Team Combinations

Total ways to select the team = Ways(Case 1) + Ways(Case 2)
Total ways = $15 + 80 = 95$ ways.

Selecting the Captain

Each selected team has $4$ members.
Any of these $4$ members can be the captain.
Number of ways to select a captain from a team: $^{4} C_{1} = 4$ .

Final Result

Total ways = (Ways to select team) $\times$ (Ways to select captain)
Total ways = $95 \times 4 = 380$
Key Takeaway: Always handle constraints first by creating cases, then apply multi-step selection logic.

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The Sigma Insight: Combinations and Selection

Solution Diagram

Analyzing the Setup

Imagine you are standing in the heart of a bustling debate club. You have a pool of

10

brilliant minds:

6

girls and

4

boys.

Your mission is to assemble an elite team of

4

members, subject to the constraint that the team must contain 'at most one boy'. This is a puzzle of logical architecture that requires careful decomposition.

Decoding the Constraint

The phrase 'at most one boy' acts as a filter, narrowing down the possibilities into two distinct, manageable realities. We cannot have two, three, or four boys.

We are left with two mutually exclusive scenarios: Case 1: $0$ boys and $4$ girls. Case 2: Exactly

1

boy and

3

girls.

By separating the problem into these two silos, we eliminate the risk of overlapping counts.

The Selection Process

Let us tackle Case 1:

0

boys and

4

girls. We need to choose

4

girls from the

6

available.

The number of ways to do this is given by the combination formula

^{6} C_{4}

. Using the symmetry property

^{n} C_{r} =^{n} C_{n - r}

, we know that

^{6} C_{4}

is identical to

^{6} C_{2}

^{6} C_{2} = \frac{6 \times 5}{2 \times 1} = 15 ways

Now, for Case 2:

1

boy and

3

girls. We select

1

boy from

4

and

3

girls from

6

^{4} C_{1} = 4

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

Since these selections happen together to form one team, we multiply them:

4 \times 20 = 80

ways. Adding these two cases together, we find the total number of ways to form the team:

15 + 80 = 95 ways

The Final Twist

The JEE examiner tests your attention to detail. We have formed the team, but we must still appoint a captain from the

4

members.

For every one of those

95

teams, any of the

4

members could be the captain. This is a multi-step selection process where we multiply the team combinations by the possible choices for the captain:

95 \times 4 = 380

The beauty of this problem lies in its structure: define your cases, calculate the combinations, and never forget the final assignment step. The final answer is 380.

Common Traps & FAQs

Why do we add the results of Case 1 and Case 2 instead of multiplying them?

Great question! We add them because these cases are mutually exclusive. A team cannot simultaneously have zero boys and exactly one boy. Since these are distinct, non-overlapping scenarios that both satisfy the condition, we sum the number of ways to get the total count.

Why do we multiply by 4 at the end? Isn't the team selection the final step?

The team selection is only the first part of the process. The problem explicitly states that we must also select a captain from the

4

chosen members. Since each unique team of

4

offers

4

distinct possibilities for the captain, we use the multiplication principle to account for this secondary selection step.

Could I have used the complement method (Total ways - ways with >1 boy)?

You absolutely could! However, in this specific problem, the direct method is faster. The complement would require you to calculate teams with

2

3

, and

4

boys and subtract them from the total possible teams. Both methods will yield

380

, but the direct method is less prone to calculation errors here.

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Visualized Solution

The Starting Pool

Defining 'At Most One Boy'

Case 1: Selecting $4$ Girls

Calculating $^{6} C_{4}$

Case 2: $1$ Boy and $3$ Girls

Calculating $^{4} C_{1} \times^{6} C_{3}$

Total Team Combinations

Selecting the Captain

Final Result

The Sigma Insight: Combinations and Selection

Analyzing the Setup

Decoding the Constraint

The Selection Process

The Final Twist

Similar Questions

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 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 :

There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is _____

List-I

List-II

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 ___

A committee of 12 is to be formed from 9 women and 8 men. In how many ways this can be done if at least five women have to be included in a committee? In how many of these committees (a) The women are in majority? (b) The men are in majority?

Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to:

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 committee of 11 members is to be formed from 8 males and 5 females. If $m$ is the number of ways the committee is formed with at least 6 males and $n$ is the number of ways the committee is formed with at least 3 females, then :

Select Answer:

Visualized Solution

The Starting Pool

Defining 'At Most One Boy'

Case 1: Selecting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4 Girls

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 6C4​

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1 Boy and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3 Girls

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4C1​×6C3​

Total Team Combinations

Selecting the Captain

Final Result

The Sigma Insight: Combinations and Selection

Analyzing the Setup

Decoding the Constraint

The Selection Process

The Final Twist

Similar Questions

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 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 :

There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is _____

List-I

List-II

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 ___

A committee of 12 is to be formed from 9 women and 8 men. In how many ways this can be done if at least five women have to be included in a committee? In how many of these committees (a) The women are in majority? (b) The men are in majority?

Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to:

Case 1: Selecting $4$ Girls

Calculating $^{6} C_{4}$

Case 2: $1$ Boy and $3$ Girls

Calculating $^{4} C_{1} \times^{6} C_{3}$