Misc 3 - Chapter 6 Class 11 Permutations and Combinations

Last updated at April 16, 2024 by Teachoo

Transcript

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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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