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Misc 3 - A committee of 7 has to be formed from 9 boys and 4 girls

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Misc 3 - Chapter 7 Class 11 Permutations and Combinations - Part 2

Misc 3 - Chapter 7 Class 11 Permutations and Combinations - Part 3 Misc 3 - Chapter 7 Class 11 Permutations and Combinations - Part 4 Misc 3 - Chapter 7 Class 11 Permutations and Combinations - Part 5

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Misc 3 - Chapter 7 Class 11 Permutations and Combinations - Part 6 Misc 3 - Chapter 7 Class 11 Permutations and Combinations - Part 7 Misc 3 - Chapter 7 Class 11 Permutations and Combinations - Part 8 Misc 3 - Chapter 7 Class 11 Permutations and Combinations - Part 9 Misc 3 - Chapter 7 Class 11 Permutations and Combinations - Part 10 Misc 3 - Chapter 7 Class 11 Permutations and Combinations - Part 11

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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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.