Check sibling questions

Example 21 - A group consists of 4 girls and 7 boys. In how many

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Example 21 - Chapter 7 Class 11 Permutations and Combinations - Part 2

Example 21 - Chapter 7 Class 11 Permutations and Combinations - Part 3
Example 21 - Chapter 7 Class 11 Permutations and Combinations - Part 4
Example 21 - Chapter 7 Class 11 Permutations and Combinations - Part 5
Example 21 - Chapter 7 Class 11 Permutations and Combinations - Part 6
Example 21 - Chapter 7 Class 11 Permutations and Combinations - Part 7

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Example 21 - Chapter 7 Class 11 Permutations and Combinations - Part 8

Example 21 - Chapter 7 Class 11 Permutations and Combinations - Part 9
Example 21 - Chapter 7 Class 11 Permutations and Combinations - Part 10
Example 21 - Chapter 7 Class 11 Permutations and Combinations - Part 11

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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 We have to calculate all these combinations separately and then add it Option 1 - 1 boy & 4 girls Number of ways selecting 1 boy & 4 girls = 7C1 × 4C4 = 7!/1!(7 − 1)! × 4!/4!(4 − 4)! = 7!/(1! 6!) × 4!/(4! 0!) = (7 × 6!)/6! × 4!/4! = 7 × 1 = 7 Option 2 – 2 boys & 3 girls Number of ways selecting 2 boys & 3 girls = 7C2 × 4C3 = 7!/2!(7 − 2)! × 4!/3!(4 − 3)! = 7!/(2! 5!) × 4!/(3! 1!) = (7 × 6 × 5!)/(2! 5!) × (4 × 3!)/3! = 21 × 4 = 84 Option 3 – 3 boys & 2 girls Number 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! 4!) × (4 × 3 × 2!)/(2! 2!) = 7 × 5 × 2 × 3 = 210 Option 4 – 4 boys & 1 girl Number of ways selecting 4 boys & 1 girl = 7C4 × 4C1 = 7!/4!(7 − 4)! × 4!/1!(4 − 1)! = 7!/(4! 3!) × 4!/(1! 3!) = (7 × 6 × 5 × 4!)/(4! 3!) × (4 × 3!)/3! = 7 × 5 × 4 = 140 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. We have to calculate all these combinations separately and then add it Option 1 – 3 girls and 2 boys Number of ways selecting 3 girls and 2 boys = 4C3 × 7C2 = 4!/3!(4 − 3)! "×" 7!/2!(7 − 2)! = 4!/(3! 1!) × 7!/(2! 5!) = (4 × 3!)/3! × (7 × 6 × 5!)/(2! × 5!) = 4 × 7 × 3 = 84 Option 2 – 4 girls and 1 boy Number of ways selecting 4 girls and 1 boy = 4C4 × 7C1 = 4!/4!(4 − 4)! "×" 7!/1!(7 − 1)! = 4!/(4! 0!) × 7!/(1! 6!) = 4!/4! × (7 × 6!)/6! = 1 × 7 = 7 Hence, Total number of ways = 84 + 7 = 91

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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.