Question

Question 3 A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:

(i) exactly 3 girls? (ii) atleast 3 girls? (iii) atmost 3 girls?

Class X1 - Maths -Permutations and Combinations Page 156

Answer 1

(i) out of 7 persons , we have to select exactly 3 girls . hence, it's clear that remaining 4 persons will be boys .

now,
number of ways taken 4 boys from 9 boys = 9C4
and number of ways taken 3 girls from 4 girls = 4C3

hence, by fundamental principle of counting ,
total number of ways for making the committee = 9C4 × 4C3
= 9!/5!.4! × 4!/3!
= 9×8×7×6/3×2×1
= 9×8×7
= 504 ways .


(ii) in committee , at least three girls .
there are two possibilities :
1st :- when take 3 girls in committee
then, number of ways taken (7-3) boys from 9 boys = 9C4
number of ways taken 3 girls from 4 girls = 4C3
hence, total number of ways in this case = 9C4 × 4C3 =504 ways

2nd case:- maximum 4 girls can be taken .
so, let we take 4 girls in committee ,
then,
number of ways taken(7-4) boys from 9 boys = 9C3
and number of ways 4 girls from 4 girls = 4C4

hence by Fundamental principle of counting,
total number of ways = 9C3 × 4C4
= 9×8×7/6 = 84 ways .

hence, by fundamental principle of addition , total number of ways = 504 + 84 = 588 ways .


(iii) here, we have to select at most 3 girls .
there are four possibilities :

1st case :- when take 3 girls.
number of ways taken 4 boys from 9 boys = 9C4
number of ways taken 3 girls from 4 girls = 4C3
hence, total number of ways= 9C4 × 4C3

2nd case:- when take 2 girls
number of ways taken 5boys from 9boys = 9C5
number of ways taken 2girls from 4girls = 4C2
hence, total number of ways= 9C5 × 4C2


3rd case:- when take 1 girls
number of ways 6boys taken from 9boys = 9C6
number of ways taken 1girls from 4 girls = 4C1
hence, total number of ways= 9C6 × 4C1

4th case:- when take 0 girls
number of ways taken 7boys from 9 boys = 9C7
number of ways taken 0girls from 4 girls = 4C0
hence, total number of ways= 9C7×4C0

hence, total number of ways = (9C4 × 4C3 ) + ( 9C5 × 4C2) + ( 9C6 × 4C1) + ( 9C7 × 4C0)
= 504 + 756 + 336 + 36
= 1632 ways