JEE Advanced 1994

LEVELJEE Main

Animated Solution for Mathematics - Permutations and Combinations: 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?

Progressive Hints

Hint 1

The phrase 'at least' is a siren song in combinatorics. It screams that you cannot solve this in one go. You must partition the problem into mutually exclusive scenarios. Visualize the committee as a bucket; how many ways can you fill it while keeping the 'women' count at 5, 6, 7, 8, or 9?

Hint 2

Remember your fundamental tools: the combination formula

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

. Since you are selecting people for a committee where order doesn't matter, this is your primary weapon. Use the Multiplication Principle for selecting within a case and the Addition Principle for combining the cases.

Hint 3

Start your engine with the base case: 5 women and 7 men. The number of ways is

^{9} C_{5} \times^{8} C_{7}

. Calculate this, then systematically increment the number of women while decrementing the number of men until you hit the limit of available women.

Visualized Solution

Problem Setup and Constraints

Total Women available: $9$
Total Men available: $8$
Committee size required: $12$
Constraint: At least $5$ women must be included.

Case 1: Minimum Women

Case 1: $5$ Women and $7$ Men

Calculating Case 1

Ways = $^{9} C_{5} \times^{8} C_{7}$
Ways = $126 \times 8$
Ways = $1008$

Case 2: Increasing Women to $6$

Case 2: $6$ Women and $6$ Men

Calculating Case 2

Ways = $^{9} C_{6} \times^{8} C_{6}$
Ways = $84 \times 28$
Ways = $2352$

Case 3: $7$ Women

Case 3: $7$ Women and $5$ Men
Ways = $^{9} C_{7} \times^{8} C_{5} = 36 \times 56 = 2016$

Case 4: $8$ Women

Case 4: $8$ Women and $4$ Men
Ways = $^{9} C_{8} \times^{8} C_{4} = 9 \times 70 = 630$

Case 5: Maximum Women ( $9$ )

Case 5: $9$ Women and $3$ Men
Ways = $^{9} C_{9} \times^{8} C_{3} = 1 \times 56 = 56$

Total Number of Committees

Total Ways = Sum of all mutually exclusive cases
Total = $1008 + 2352 + 2016 + 630 + 56$
Total = $6062$

Part (a): Condition for Women Majority

Condition for Women Majority: Women $>$ Men
In a committee of $12$ , Women $\geq 7$
Valid Cases: Case $3$ , Case $4$ , Case $5$

Part (a): Calculating Women Majority

Women Majority Ways = $2016 + 630 + 56$
Women Majority Ways = $2702$

Part (b): Men in Majority

Condition for Men Majority: Men $>$ Women
In a committee of $12$ , Men $\geq 7$
Valid Case: Case $1$ ( $7$ Men, $5$ Women)
Men Majority Ways = $1008$

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The Sigma Insight: Combinations and Selection

Solution Diagram

Analyzing the Setup

To form a committee of 12 members from a group of 9 women and 8 men with the constraint of having at least 5 women, we must partition the problem into mutually exclusive scenarios. This approach transforms a complex constraint into a series of manageable, additive cases.

The combination formula used throughout this process is defined as:

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

Case-by-Case Analysis

We identify five distinct scenarios based on the number of women (

W

) and men (

M

) selected to reach a total of 12 members:

Case 1: 5 Women and 7 Men Case 2: 6 Women and 6 Men Case 3: 7 Women and 5 Men Case 4: 8 Women and 4 Men Case 5:* 9 Women and 3 Men

The Calculation Engine

We calculate the number of ways for each case by multiplying the combinations of women and men:

Case 1 (5W, 7M):

^{9} C_{5} \times^{8} C_{7} = 126 \times 8 = 1008

Case 2 (6W, 6M):

^{9} C_{6} \times^{8} C_{6} = 84 \times 28 = 2352

Case 3 (7W, 5M):

^{9} C_{7} \times^{8} C_{5} = 36 \times 56 = 2016

Case 4 (8W, 4M):

^{9} C_{8} \times^{8} C_{4} = 9 \times 70 = 630

Case 5 (9W, 3M):

^{9} C_{9} \times^{8} C_{3} = 1 \times 56 = 56

The Grand Total

By applying the Addition Principle to these mutually exclusive cases, we find the total number of ways to form the committee:

1008 + 2352 + 2016 + 630 + 56 = 6062

Majority Analysis

Part (a): Women in the majority. This requires more than 6 women (i.e., 7, 8, or 9 women). Summing the results from Case 3, Case 4, and Case 5:

2016 + 630 + 56 = 2702

Part (b): Men in the majority. This requires more than 6 men (i.e., 7 men). Only Case 1 satisfies this condition:

1008

Common Traps & FAQs

Why can't I just select 5 women first and then pick the remaining 7 members from the remaining 12 people?

This is a classic trap! If you pick 5 women first and then 7 from the remaining 12, you are implicitly ordering the selection process. This leads to massive overcounting because the same committee could be formed in different sequences of selection. By using mutually exclusive cases, we ensure every unique committee is counted exactly once.

Why do we add the cases instead of multiplying them?

We add the cases because they are mutually exclusive. A committee cannot simultaneously have exactly 5 women and exactly 6 women. Since these scenarios are distinct, the Fundamental Principle of Counting dictates that we sum the number of ways for each scenario to find the total.

Is there a faster way to do this using subtraction (total - complement)?

You could calculate the total number of ways to form a committee of 12 from 17 people (

^{17} C_{12}

) and subtract the cases where there are fewer than 5 women (0, 1, 2, 3, or 4 women). However, in this specific problem, calculating the 5 cases directly is actually faster and less prone to arithmetic errors than calculating the 5 complement cases!

Visualized Solution

Problem Setup and Constraints

Case 1: Minimum Women

Calculating Case 1

Case 2: Increasing Women to $6$

Calculating Case 2

Case 3: $7$ Women

Case 4: $8$ Women

Case 5: Maximum Women ( $9$ )

Total Number of Committees

Part (a): Condition for Women Majority

Part (a): Calculating Women Majority

Part (b): Men in Majority

The Sigma Insight: Combinations and Selection

Analyzing the Setup

Case-by-Case Analysis

The Calculation Engine

The Grand Total

Majority Analysis

Similar Questions

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 :

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:

List-I

List-II

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 _____

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 ______.

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:

Eighteen guests have to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on the other side. Determine the number of ways in which the sitting arrangements can be made.

7 relatives of a man comprises 4 ladies and 3 gentlemen; his wife has also 7 relatives; 3 of them are ladies and 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of man's relatives and 3 of the wife's relatives?

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 :

Visualized Solution

Problem Setup and Constraints

Case 1: Minimum Women

Calculating Case 1

Case 2: Increasing Women to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 6

Calculating Case 2

Case 3: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 7 Women

Case 4: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 8 Women

Case 5: Maximum Women ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 9)

Total Number of Committees

Part (a): Condition for Women Majority

Part (a): Calculating Women Majority

Part (b): Men in Majority

The Sigma Insight: Combinations and Selection

Analyzing the Setup

Case-by-Case Analysis

The Calculation Engine

The Grand Total

Majority Analysis

Similar Questions

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:

List-I

List-II

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 _____

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:

Eighteen guests have to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on the other side. Determine the number of ways in which the sitting arrangements can be made.

7 relatives of a man comprises 4 ladies and 3 gentlemen; his wife has also 7 relatives; 3 of them are ladies and 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of man's relatives and 3 of the wife's relatives?

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 :

Case 2: Increasing Women to $6$

Case 3: $7$ Women

Case 4: $8$ Women

Case 5: Maximum Women ( $9$ )