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

  1. Chapter 7 Class 11 Permutations and Combinations
  2. Serial order wise
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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 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 Option 1 3 girls and 4 boys Total 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 Option 2 4 girls & 3 boys Total ways = 4C4 × 9C3 = 4!/(4!(4 − 4)!) × 9!/3!(9 − 3)! = 4!/4!0! × 9!/3!(6)! = 1 × 9!/3!(6)! = (9 × 8 × 7 × 6!)/((3 × 2 × 1) × (6)!) = 84 Hence, total 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 Option 1 0 girls & 7 boys Number of ways selecting no girls & 7 boys = 9C7 × 4C0 = 9!/7!(9 − 7)! × 4!/0!(4 − 0)! = 9!/(7! × 2!) × 4!/0!4! = (9 × 8 × 7!)/(7! × 2 × 1) × 4!/4! = (9 × 8)/2 × 1 = 36 × 1 = 36 Option 2 1 girl & 6 boys Number of ways selecting 6 boys & one girls = 9C6 × 4C1 = 9!/6!(9 − 6)! × 4!/1!(4 − 1)! = 9!/6!3! × 4!/1!3! = (9 × 8 × 7 × 6!)/(6! × 3!) × (4 × 3!)/3! = (9 × 8 × 7)/3! × 4 = (9 × 8 × 7 × 4)/(3 × 2) = 336 Option 3 2 girls & 5 boys Number of ways selecting no girls & 7 boys = 9C5 × 4C2 = 9!/5!(9 − 5)! × 4!/2!(4 − 2)! = 9!/(5! × 4!) × 4!/2!2! = (9 × 8 × 7 × 6 × 5!)/5!4! × 4!/(2 × 1 × 2 × 1) = (9 × 8 × 7 × 6 × 5! × 4!)/(2 × 2 × 1 × 1 × 5! × 4!) = (9 × 8 × 7 × 6)/(2 × 2) = 756 Option 4 3 girls & 4 boys Number of ways selecting 3 girls & 4 boys = 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 Total ways = 36 + 336 +756 + 504 = 1632 ways

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.