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Animated Solution for Mathematics - Permutations and Combinations: 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 :

Q: Is it possible to have a committee with fewer than $6$ males?

Great question! Let's look at the constraints. We need $11$ members. We only have $5$ females. If we picked only $5$ males, we would have $5+5=10$ members, which is not enough. Thus, we *must* have at least $6$ males. The constraint is naturally satisfied!

Q: Why is $m=n$?

Because the total committee size is $11$ and the total pool is $13$, the number of males and females in any committee are linked. If you have $x$ males, you must have $11-x$ females. The conditions $\ge 6$ males and $\ge 3$ females turn out to be equivalent because of this fixed sum.

Select Answer:

Visualized Solution

Problem Setup

Total available: $8$ Males + $5$ Females = $13$ people.
Committee size to be formed: $11$ members.

Condition for $m$ : $\geq 6$ Males

$m$ is the number of ways to form the committee with at least $6$ males.
Possible cases: $6$ M + $5$ F, $7$ M + $4$ F, $8$ M + $3$ F.

Case 1: $6$ Males and $5$ Females

Select $6$ Males from $8$ : $(6 8) = (2 8) = 28$ .
Select $5$ Females from $5$ : $(5 5) = 1$ .
Total ways for Case 1: $28 \times 1 = 28$ .

Case 2: $7$ Males and $4$ Females

Select $7$ Males from $8$ : $(7 8) = 8$ .
Select $4$ Females from $5$ : $(4 5) = 5$ .
Total ways for Case 2: $8 \times 5 = 40$ .

Case 3: $8$ Males and $3$ Females

Select $8$ Males from $8$ : $(8 8) = 1$ .
Select $3$ Females from $5$ : $(3 5) = 10$ .
Total ways for Case 3: $1 \times 10 = 10$ .

Total Ways for $m$

Summing all cases for $m$ :
$m = 28 + 40 + 10 = 78$

Condition for $n$ : $\geq 3$ Females

$n$ is the number of ways to form the committee with at least $3$ females.
Possible cases: $3$ F + $8$ M, $4$ F + $7$ M, $5$ F + $6$ M.

Logical Equivalence: $m = n$

The cases for $n$ ( $3$ F+ $8$ M, $4$ F+ $7$ M, $5$ F+ $6$ M) are exactly the same as the cases for $m$ .
Therefore, $n = m = 78$ .

The Ultimate Shortcut

Max females = $5$ . To form an $11$ -member committee, min males = $11 - 5 = 6$ .
Thus, every valid committee naturally satisfies both conditions!
Total ways = $(11 13) = (2 13) = \frac{13 \times 12}{2} = 78$ .

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The Sigma Insight: Combinations and Selection

Solution Diagram

Analyzing the Setup

We are tasked with forming a committee of

11

members from a group of

13

people, consisting of

8

males and

5

females.

To solve this, we define

m

as the number of ways to form a committee with at least

6

males, and

n

as the number of ways to form a committee with at least

3

females.

The Brute Force Approach

Let us calculate

m

by considering the mutually exclusive cases for the number of males chosen:

If we choose

6

males, we must choose

5

females:

(6 8) \times (5 5) = 28 \times 1 = 28

If we choose

7

males, we must choose

4

females:

(7 8) \times (4 5) = 8 \times 5 = 40

If we choose

8

males, we must choose

3

females:

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

Summing these values, we find

m = 28 + 40 + 10 = 78

The Hidden Symmetry

Now, consider the constraints for

n

, which requires at least

3

females.

Note that we are selecting

11

members out of

13

, which is equivalent to excluding

2

people.

If we select the maximum possible number of females (

5

), we are still forced to select

6

males to reach the committee size of

11

Conversely, if we select the maximum possible number of males (

8

), we are forced to select

3

females to reach the committee size of

11

The Master's Shortcut

Because of the committee size constraint, every valid committee of

11

people automatically contains at least

6

males and at least

3

females.

The conditions for

m

and

n

are satisfied by every possible combination of

11

people chosen from the

13

available.

Therefore, both

m

and

n

are equal to the total number of ways to choose

11

people from

13

m = n = (11 13) = (2 13)

Calculating this value:

(2 13) = \frac{13 \times 12}{2} = 78

The final result is $m = n = 78$ .

Common Traps & FAQs

Why do we have to split into cases for

m

Splitting into cases is a robust, fundamental method. It ensures you don't miss any combinations. While there is a shortcut, understanding the case-by-case breakdown is essential for building your combinatorial intuition.

Is it possible to have a committee with fewer than

6

males?

Great question! Let's look at the constraints. We need

11

members. We only have

5

females. If we picked only

5

males, we would have

5 + 5 = 10

members, which is not enough. Thus, we *must* have at least

6

males. The constraint is naturally satisfied!

Why is

m = n

Because the total committee size is

11

and the total pool is

13

, the number of males and females in any committee are linked. If you have

x

males, you must have

11 - x

females. The conditions

\geq 6

males and

\geq 3

females turn out to be equivalent because of this fixed sum.

Animated Solution for Mathematics - Permutations and Combinations: 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

Problem Setup

Condition for $m$ : $\geq 6$ Males

Case 1: $6$ Males and $5$ Females

Case 2: $7$ Males and $4$ Females

Case 3: $8$ Males and $3$ Females

Total Ways for $m$

Condition for $n$ : $\geq 3$ Females

Logical Equivalence: $m = n$

The Ultimate Shortcut

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Brute Force Approach

The Hidden Symmetry

The Master's Shortcut

Similar Questions

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?

List-I

List-II

A scientific committee is to formed from 6 Indians and 8 foreigners, which includes at least 2 Indians and double the number of foreigners as Indians. Then the number of ways, the committee can be formed is:

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

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 _____

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 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 ______.

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:

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:

Select Answer:

Visualized Solution

Problem Setup

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Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 6 Males and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5 Females

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 7 Males and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4 Females

Case 3: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 8 Males and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3 Females

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

Condition for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ≥3 Females

Logical Equivalence: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m=n

The Ultimate Shortcut

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Brute Force Approach

The Hidden Symmetry

The Master's Shortcut

Similar Questions

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?

List-I

List-II

A scientific committee is to formed from 6 Indians and 8 foreigners, which includes at least 2 Indians and double the number of foreigners as Indians. Then the number of ways, the committee can be formed is:

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

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 _____

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 :

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:

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:

Condition for $m$ : $\geq 6$ Males

Case 1: $6$ Males and $5$ Females

Case 2: $7$ Males and $4$ Females

Case 3: $8$ Males and $3$ Females

Total Ways for $m$

Condition for $n$ : $\geq 3$ Females

Logical Equivalence: $m = n$