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Last updated at Jan. 30, 2020 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 π!/π1!π2!π3! 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 β 1) = 11 We need to arrange letters I, N, D, E, E, N, D, E, N, C, E Here, we have 4E, 3N,2D Since letters are repeating since we use this formula Number of arrangements = π!/π1!π2!π3! Total letters to arrange = 11 So, n = 11 Since, 4E, 3N,2D p1 = 4 , p2 = 3 , p3 = 2 Number of arrangements = π!/π1!π2!π3! = 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 βINDEPENDENCEβ i.e. 4Eβs & Iβs They have occur together we treat them as single object we treat as a single object So our letters become We arrange them now Arranging 5 vowels Since vowels are coming together, they can be and so on In EEEEI, there are 4E Since letters are repeating, We use this formula = π!/π1!π2!π3! Total letter = n = 5 As 4E are there, p1 = 4 Total arrangements = 5!/4! Arranging remaining letters Numbers we need to arrange = 7 + 1 = 8 Here are 3N, 2D Since letter are repeating, We use this formula = π!/π1!π2!π3! Total letters = n = 8 As 3N, 2D p1 = 3 , p2 = 2 Total arrangements = 8!/3!2! Hence, Required number of arrangements = 8!/(3! 2!) Γ 5!/4! = ((8 Γ 7 Γ 6 Γ 5 Γ 4 Γ 3!) Γ (5 Γ 4!))/(3! Γ 4! Γ 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 β Number of arrangements when all vowels occur together = 1663200 β 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 use the formula π!/π1!π2!π3! Here, Total letters = n = 10 Since 2D, 4E, 3N p1 = 2, p2 = 4, p3 = 3 Required number of arrangement = 10!/2!4!3! = 12600

Chapter 7 Class 11 Permutations and Combinations

Class 11

Important Questions 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.