# 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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