Ex 6.3, 10 - In how many distinct permutations in MISSISSIPPI - Ex 6.3

part 2 - Ex 6.3,10 - Ex 6.3 - Serial order wise - Chapter 6 Class 11 Permutations and Combinations
Ex 6.3.jpg part 4 - Ex 6.3,10 - Ex 6.3 - Serial order wise - Chapter 6 Class 11 Permutations and Combinations

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Ex 6.3, 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 Total Permutations In MISSISSIPPI there are 4I, 4S, 2P and 1M Since letters are repeating, we will use the formula = 𝑛!/𝑝1!𝑝2!𝑝3! Total number of alphabet = 11 Hence n = 11, Also, there are 4I, 4S, 2P p1 = 4, p2 = 4, p3 = 2 Hence, Total number of permutations = 𝑛!/𝑝1!𝑝2!𝑝3! = 11!/(4! 4! 2!) = (11 × 10 × 9 × 8 × 7 × 6 × 5 × 4!)/((4 × 3 × 2 × 1) (4!)×(2 × 1)) = 34650 Total permutations of I coming together 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 ∴ n = 8 Since there are 4 S & 2 P p1 = 4, p2 = 2, Number of permutation with 4I together = 𝑛!/𝑝1!𝑝2! = 8!/(4! 2!) = 840 Now, Total number of permutation of 4I not coming together = Total permutation – total permutation of I coming together = 34650 – 840 = 33810

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