A committee of 12 is to be formed from...

MathematicsCombinations and SelectionJEE Advanced 1994Easy

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?

Visualized Solution (Hindi)

Problem Setup and Constraints

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

Identifying the Possible Cases

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

Calculating Case 1 and Case 2

Case 1: $^{9} C_{5} \times^{8} C_{7} = 126 \times 8 = 1008$
Case 2: $^{9} C_{6} \times^{8} C_{6} = 84 \times 28 = 2352$

Calculating Case 3, 4, and 5

Case 3: $^{9} C_{7} \times^{8} C_{5} = 36 \times 56 = 2016$
Case 4: $^{9} C_{8} \times^{8} C_{4} = 9 \times 70 = 630$
Case 5: $^{9} C_{9} \times^{8} C_{3} = 1 \times 56 = 56$

Total Number of Committees

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

Women in Majority

Women in majority if Women $>$ Men
In a committee of $12$ , this means Women $\geq 7$
Cases: $7 W, 8 W, 9 W$
Ways = $2016 + 630 + 56 = 2702$

Men in Majority

Men in majority if Men $>$ Women
In a committee of $12$ , this means Men $\geq 7$
Valid Case: $7$ Men and $5$ Women (Case 1)
Ways = $1008$

Summary and Key Takeaways

Total Ways: $6062$
Women Majority: $2702$ ways
Men Majority: $1008$ ways
Key Principle: Always list mutually exclusive cases first when dealing with 'at least' or 'at most' constraints.

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