PRACTICE WORD PROBLEMS ON COMBINATIONS WITH SOLUTIONS

Problem 1 :

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

(i) exactly 3 women?

(ii) at least 3 women?

(iii) at most 3 women?

Solution :

Number of men  =  8

Number of women  =  4

(i) exactly 3 women?

Number of ways  =  ⁴C₃ ⋅  ⁸C₄

=  4 (70)

=  280

(ii) at least 3 women?

Number of ways  =  (⁴C₃ ⋅  ⁸C₄) +  (⁴C₄ ⋅  ⁸C₃)

  =  4(70) + 1(56)

  =  280 + 56

  =  336

(iii) at most 3 women?

Number of ways 

  =  (⁴C₀ ⋅  ⁸C₇) +  (⁴C₁ ⋅  ⁸C₆) +  (⁴C₂ ⋅  ⁸C₅)  +  (⁴C₃ ⋅  ⁸C₄)

  =  8 + 4(28) + 6(56) + 4(70)

  =   8 + 112 + 336 + 280

  =  736

Problem 2 :

7 relatives of a man comprises 4 ladies and 3 gentlemen, his wife also has 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 relative and 3 of the wife’ s relatives?

Solution :

(i) 3 ladies from husband’s side and 3 gentlemen from wife’s side.

No. of ways in this case

  = ⁴C₃ ⋅ ⁴C₃ = 4 ⋅ 4 = 16

(ii) 3 gentlemen from husband’s side and 3 ladies from wife’s side.

No. of ways in this case = ³C₃ ⋅ ³C₃ = 1 ⋅ 1 = 1

(iii) 2 ladies and one gentleman from husband’s side and lady and 2 gentlemen from wife’s side.

No. of ways in this case

= (⁴C₄ ⋅ ³C₁) ⋅ (³C₁ ⋅ ⁴C₂) = 6 ⋅ 3 ⋅ 3 ⋅ 6 = 324

(iv) One lady and 2 gentlemen from husband’s side and 2 ladies and one gentlemen from wife’s side.

No. of ways in this case

  =  (⁴C₁ ⋅ ³C₂) ⋅ (³C₂ ⋅ ⁴C₁) = 4 ⋅ 3 ⋅ 3 ⋅ 4 = 144

Hence the total no. of ways are

=  16 + 1 + 324 + 144 = 485 ways 

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