Question

A committee of 10 persons is to be formed from a group of 10 women and 8 men. How many possible committees will have at least 5 women? How many possible committees will have men in majority?

Answer 1

There are 10 women and 8 men.

A committee of 10 persons is to be formed. If at least 5 women have been included in a committee, then the possible selection are :

5 women and 5 men, 6 women and 4 men,

7 women and 3 men, 8 women and 2 men,

9 women and 1 man and all the 10 women and no man.

∴ the number of ways of forming committees such that at least five women are included

= ¹⁰C₅ × ⁸C₅ + ¹⁰C₆ × ⁸C₄ + ¹⁰C₇ × ⁸C₃ + ¹⁰C₈ × ⁸C₂ + ¹⁰C₉ × ⁸C₁ + ¹⁰C₁₀ × ⁸C₀

= `(10!)/(5!5!) xx (8!)/(5!3!) + (10!)/(6!4!) xx (8!)/(4!4!) + (10!)/(7!3!) xx (8!)/(3!5!) + (10!)/(8!2!) xx (8!)/(2!6!) + (10!)/(9!1!) xx (8!)/(1!7!) + (10!)/(10!0!) xx (8!)/(0!8!)`

= `((10 xx 9 xx 8 xx 7 xx 6)/(5 xx 4 xx 3 xx 2 xx 1) xx (8 xx 7 xx 6)/(1 xx 2 xx 3)) + ((10 xx 9 xx 8 xx 7)/(1 xx 2 xx 3 xx 4) xx (8 xx 7 xx 6 xx 5)/(1 xx 2 xx 3 xx 4)) + ((10 xx 9 xx 8)/(1 xx 2 xx 3) xx (8 xx 7 xx 6)/(1 xx 2 xx 3)) + ((10 xx 9)/(1 xx 2) xx (8 xx 7)/(1 xx 2)) + (10 xx 8) + (1 xx 1)`

= (252 × 56) + (210 × 70) + (120 × 56) + (45 × 28) + 80 + 1

= 14112 + 14700 + 6720 + 1260 + 80 + 1

= 36873

Men in majority :

In a committee of 10 persons if men are in majority, then they are 6 or 7 or 8 in numbers.

∴ the number of committees in which men are in majority

= ⁸C₆ × ¹⁰C₄ + ⁸C₇ × ¹⁰C₃ + ⁸C₈ × ¹⁰C₂ 

= `(8!)/(6!2!) xx (10!)/(4!6!) + (8!)/(7!1!) xx (10!)/(3!7!) + (8!)/(8!0!) xx (10!)/(2!8!)`

= `((8 xx 7)/(1 xx 2) xx (10 xx 9 xx 8 xx 7)/(1 xx 2 xx 3 xx 4)) + (8 xx (10 xx 9 xx 8)/(1 xx 2 xx 3)) + (1 xx (10 xx 9)/(1 xx 2))`

= (28 × 210) + (8 × 120) + 45

= 5880 + 960 + 45

= 6885