Last updated at March 8, 2017 by Teachoo

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Example 16 Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, Finding total number of arrangements In word INDEPENDENCE There are 3N, 4E, & 2D, 1I, 1P & 1C Since letters are repeating so we use this formula 𝑛!𝑝1!𝑝2!𝑝3! Total letters = 12 So, n = 12 Since, 3N, 4E, & 2D p1 = 3, p2 = 4,p3 = 2 Total arrangements = 12!3!4!2! = 1663200 Example 16 Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, • do the words start with P If the word start with P We need to arrange (12 – 1) = 11 We need to arrange letters I,N,D,E,E,N,D,E,N,C,E Here, 4E, 3N,2D Since letters are repeating since we use this formula Number of arrangements = 𝑛!𝑝1!𝑝2!𝑝3! Total letters to arrange = 11 So, n = 11 Since, 4E, 3N,2D p1 = 4 , p2 = 3 , p3 = 2 Number of arrangements = 𝑛!𝑝1!𝑝2!𝑝3! = 11!4!3!2! = 138600 Example 16 Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, do the words start with P If the word start with P We need to arrange (12 – 1) = 11 We need to arrange letters I,N,D,E,E,N,D,E,N,C,E Here, 4E, 3N,2D Since letters are repeating since we use this formula Number of arrangements = 𝑛!/𝑝1!𝑝2!𝑝3! Total letters to arrange = 11 So, n = 11 Since, 4E, 3N,2D p1 = 4 , p2 = 3 , p3 = 2 Number of arrangements = 𝑛!/𝑝1!𝑝2!𝑝3! = 11!/4!3!2! = 138600 Example 16 Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, (ii) do all the vowels always occur together There are 5 vowels in the given word “INDEPENDENCE” i.e. 4E’s & I’s They have occur together we treat them as single object we treat as a single object So our letters become We arrange them now Hence the required number of arrangement = 8!/3!2! × 5!/4! = ((8 × 7 × 6 × 5 × 4 × 3!) ×(5 × 4!))/(3! × 4! × 2) = 16800 Example 16 Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, (iii) do the vowels never occur together Number of arrangements where vowel never occur together = Total number of arrangement – Number of arrangements when all the vowels occur together = 1663200 – 16800 = 1646400 Example 16 Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, (iv) do the words begin with I and end in P? Lets fix I and P at Extreme ends Since letters are repeating, Hence we using this formula 𝑛!/𝑝1!𝑝2!𝑝3! Here Total letters = n = 10 Since 2D, 4E, 3N p1 = 2, p2 = 4, p3 = 3 Required number of arrangement = 10!/2!4!3! = 12600

Ex 7.1,6
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Ex 7.1,4 Important

Example 13 Important

Example 16 Important You are here

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

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.