# Example 21

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 21 A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has no girl ? Total number of ways = 4C0 × 7C5 = 4!/0!(4 − 0)! × 7!/5!(7 − 5)! = 4!/(1 × (4)!) ×7!/5!2! = 1 × (7 × 6 × 5!)/(5! × 2 × 1) = (7 × 6)/2 = 21 Example, 21 A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has (ii) at least one boy and one girl ? A group giving at least one boy & one girl will consist of Option 1 1 boys, and 4 girls Option 2 2 boys, and 3 girls Option 3 3 boys, and 2 girls Option 4 4 boys, and 1 girls Option 1 If a group consist of 1 boys & 4 girls No of ways selecting 1 boys & 4 girls = 7C1 × 4C4 = 7!/1!(7 − 1)! × 4!/4!(4 − 4)! = 7!/1!6! × 4!/4!0! = 7!/1!6! × 4!/4!1 = (7 × 6!)/6! × 4!/4! = 7 × 1 = 7 Option 2 If there are 2 boys & 3 girls Number of ways selecting 2 boys and 3 girls = 7C2 × 4C3 = 7!/2!(7 − 2)! × 4!/3!(4 − 3)! = 7!/2!5! × 4!/3!1! = (7 × 6 × 5!)/(2 × 1 × 5!) × (4 × 3!)/(3! × 1) = (7 × 6)/(2 × 1) × 4/1 = 84 ways Option 3 If there are 3 boys & 2 girls No of ways selecting 3 boys & 2 girls = 7C3 × 4C2 = 7!/3!(7 − 3)! × 4!/2!(4 − 2)! = 7!/3!4! × 4!/2!2! = (7 × 6 × 5 × 4!)/(3 × 2 × 1 × 4!) × (4 × 3 × 2!)/(2! × 2 × 1) = (7 × 6 × 5)/(3 × 2 × 1) × (4 × 3)/(2 × 1) = 210 ways Option 4 If there are 4 boys & 1 girl No of ways selecting 4 boys & 1 girls = 7C4 × 4C1 = 7!/4!(7 − 4)! × 4!/1!(4 − 1)! = 7!/4!3! × 4!/1!3! = (7 × 6 × 5 × 4!)/(3 × 2 × 1 × 4!) × (4 × 3!)/(3! × 1) = (7 × 6 × 5)/(3 × 2 × 1) × (4 )/1 = 140 ways Hence, total number of ways = 7 + 84 + 210 + 140 = 441 ways Example, 21 A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has (iii) at least 3 girls ? Since, the team has to consist of at least 3 girls, the team can consist of 3 girls and 2 boys 4 girls and 1 boy. Option 1 3 girls & 2 boys No of ways selecting = 7C2 × 4C3 = 7!/2!(7 − 2)! × 4!/3!(4 − 3)! = 7!/2!5! × 4!/3!1! = (7 × 6 × 5!)/(2 × 1 × 5!) × (4 × 3!)/(3! × 1) = (7 × 6 × 4)/(2 × 1) = 84 Option 2 4 girls & 1 boys No of ways selecting = 7C1 × 4C4 = 7!/1!(7 − 1)! × 4!/4!(4 − 4)! = 7!/1!6! × 4!/4!0! = 7!/1!6! × 4!/4!1 = (7 × 6!)/6! × 4!/4! = 7 × 1 = 7 Total number of ways = 84 + 7 = 91

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Example 13 Important

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Example 21 You are here

Example 22

Example 23 Important

Example 24

Chapter 7 Class 11 Permutations and Combinations

Serial order wise

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .