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Example 16 - Find number of arrangements of INDEPENDENCE - Permutation- repeating

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Example 16 - Examples

  1. Class 11
  2. Important Question for exams Class 11
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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

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