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Ex 7.3, 10 - In how many distinct permutations in MISSISSIPPI - Permutation- repeating

  1. Class 11
  2. Important Question for exams Class 11
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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 Γ— 10 Γ— 9 Γ— 8 Γ— 7 Γ— 6 Γ— 5 Γ— 4!)/((4 Γ— 3 Γ— 2 Γ— 1)Γ—(2 Γ— 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

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