Last updated at May 29, 2018 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

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Example 7

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13 Important

Example 14

Example 15

Example 16 Important

Example 17

Example 18

Example 19 Important

Example 20

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

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.