Question

A committee of 7 peoples has to be formed from 8 men and 4 women. In how many ways can this be done when the committee consists of at most 3 women?

Answer 1

Number of men = 8

Number of women = 4

Number of peoples in the committee = 7

At most 3 women

The 7 members must contain at most 3 women,

∴ We have the following possibilities

(i) No women + 7 men

(ii) 1 women + 6 men

(iii) 2 women + men

(iv) 3 women + 4 men

Case (i): 0 women + 7 men

The number of ways of selecting 0 women from 4 women is = ⁴C₀

The number of ways of selecting 7 men from 8 men is = ⁸C₇

Total number of ways = ⁴C₀ × ⁸C₇

Case (ii): 1 women + 6 men

The number of ways of selecting 1 woman from 4 women is = ⁴C₁

The number of ways of selecting 6 men from 8 men is = ⁸C₆

Total number of ways = ⁴C₁ × ⁸C₆

Case (iii): 2 women + 5 men

The number of ways of selecting 2 women from 4 women is = ⁴C₃

The number of ways of selecting 4 men from 8 men is = ⁸C₄

∴ Total number of ways = ⁴C₃ × ⁸C₄

∴ The required number of ways of forming the committee

= ⁴C₀ × ⁸C₇ × ⁴C₁ × ⁸C₆ + ⁴C₂ × ⁸C₅ + ⁴C₃ × ⁸C₆ 

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

= `(8!)/(7 xx 1!) + (4!)/(11 xx 3!) xx (8!)/(6! xx 2!) + (4!)/(2! xx 2!) xx (8!)/(5! xx 3!) + (4!)/(3! xx 1!) xx (8!)/(4! xx 4!)`

= `(8 xx 7!)/(7!) + ( xx 3!)/(3!) xx (8 xx 7 xx 6!)/(6! xx 2!) + (4 xx 3 xx 2!)/(2! xx 2!) xx (8 xx 7 xx 6 xx 5!)/(5! xx 3!) + (4 xx 3!)/(3!) xx (8 xx 7 xx 6 xx 5 xx 4!)/(4! xx 4!)`

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

= 8 + 4 × 4 × 7 + 2 × 3 × 8 × 7 + 4 × 2 × 7 × 5

= 8 + 112 + 336 + 280

= 736