Misc 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:
exactly 3 girls?
Total number of ways = 4C3 × 9C4
= 4!/(3!(4 − 3)!) × 9!/4!(9 − 4)!
= 4!/3!1! × 9!/4!(5)!
= 9!/3!(5)! = (9 × 8 × 7 × 6 × 5!)/((3 × 2 × 1) × (5)!) = 504
Misc 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:
(ii) atleast 3 girls?
Since, the team has to consist of at least 3 girls, the team can consist of
3 girls and 4 boys
4 girls and 3 boys
We have to calculate all these combinations separately and then add it
Option 1 – 3 girls and 4 boys
Number of ways selecting 3 girls and 4 boys
= 4C3 × 9C4
= 4!/3!(4 − 3)! "×" 9!/4!(9 − 4)!
= 4!/(3! 1!) × 9!/(4! 5!)
= (4 × 3!)/3! × (9 × 8 × 7 × 6 × 5!)/(4! × 5!)
= 4 × 9 × 2 × 7
= 504
Option 2 – 4 girls and 3 boys
Number of ways selecting 4 girls and 3 boys
= 4C4 × 9C3
= 4!/4!(4 − 4)! "×" 9!/3!(9 − 3)!
= 4!/(4! 0!) × 9!/(3! 6!)
= 4!/4! × (9 × 8 × 7 × 6!)/(3! 6!)
= 1 × 3 × 4 × 7
= 84
Hence,
Total number of ways = 504 + 84
= 588 ways
Misc 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:
(iii) at most 3 girls?
Atmost means maximum,
We have to choose maximum 3 girls
Option 1 - 0 girls, 7 boys
Option 2 - 1 girls, 6 boys
Option 3 - 2 girls, 5 boys
Option 4 - 3 girls, 4 boys
We have to calculate all these combinations separately and then add it
Option 1 – 0 girls and 7 boys
Number of ways selecting 0 girls and 7 boys
= 4C0 × 9C7
= 4!/0!(4 − 0)! "×" 9!/7!(9 − 7)!
= 4!/(0! 4!) × 9!/(7! 2!)
= 1 × (9 × 8 × 7! )/(7! × 2!)
= 9 × 4
= 36
Option 2 – 1 girl and 6 boys
Number of ways selecting 1 girl and 6 boys
= 4C1 × 9C6
= 4!/1!(4 − 1)! "×" 9!/6!(9 − 6)!
= 4!/(1! 3!) × 9!/(6! 3!)
= (4 × 3!)/3! × (9 × 8 × 7 × 6!)/(6! × 3!)
= 4 × 3 × 4 × 7
= 336
Option 3 – 2 girls and 5 boys
Number of ways selecting 2 girl and 5 boys
= 4C2 × 9C5
= 4!/2!(4 − 2)! "×" 9!/5!(9 − 5)!
= 4!/(2! 2!) × 9!/(5! 4!)
= (4 × 3 × 2!)/(2! 2!) × (9 × 8 × 7 × 6 × 5!)/(5! × 4!)
= 2 × 3 × 9 × 2 × 7
= 756
Option 4 – 3 girls and 4 boys
Number of ways selecting 3 girls and 4 boys
= 4C3 × 9C4
= 4!/3!(4 − 3)! "×" 9!/4!(9 − 4)!
= 4!/(3! 1!) × 9!/(4! 5!)
= (4 × 3!)/3! × (9 × 8 × 7 × 6 × 5!)/(4! × 5!)
= 4 × 9 × 2 × 7
= 504
Hence,
Total number of ways = 36 + 336 +756 + 504
= 1632 ways

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.