Last updated at March 8, 2017 by Teachoo

Transcript

Ex7.3, 10 In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together? Total number of permutation of 4I not coming together = Total permutation – total permutation of I coming together In MISSISSIPPI there are 4I, 4S, 2P and 1M Since letter are repeating we will use the formula = 𝑛!/𝑝1!𝑝2!𝑝3! Total number of alphabet = 11 Hence n = 11, there are 4I, 4S, 2P p1 = 4, p2 = 4, p3 = 2 Hence total number of permutation = 𝑛!/𝑝1!𝑝2!𝑝3! = 11!/4!4!2! = (11 × 10 × 9 × 8 × 7 × 6 × 5 × 4!)/((4 × 3 × 2 × 1)×(2 × 1)(4!)) = 34650 Now taking 4Is as one, MISSISSIPPI Here, there are repeating letters So, we use the formula , Number of permutation = 𝑛!/𝑝1!𝑝2! Number of letters = 8 Since there are 4 S & 2p p1 = 4, p2 = 2, Number of permutation with 4I together = 𝑛!/𝑝1!𝑝2! = 8!/4!2! = 840 Total number of permutation of 4I not coming together = Total permutation – total permutation of I coming together = 34650 – 840 = 33810

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 You are here

Example 19 Important

Ex 7.4, 6 Important

Ex 7.4, 8 Important

Example 23 Important

Misc 3 Important

Misc 4 Important

misc 7 Important

Misc 10 Important

Misc 11 Important

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 provides courses for Mathematics from Class 9 to 12. You can ask questions here.