Last updated at May 29, 2018 by Teachoo

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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

Chapter 7 Class 11 Permutations and Combinations

Ex 7.1,6
Important

Ex 7.1,4 Important

Example 13 Important

Example 16 Important

Ex 7.3,3 Important

Ex 7.3,6 Important

Ex 7.3,10 Important

Example 19 Important

Ex 7.4, 6 Important

Ex 7.4, 8 Important

Example 23 Important

Misc 3 Important You are here

Misc 4 Important

misc 7 Important

Misc 10 Important

Misc 11 Important

Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.