How many permutations of the letters of the word MISSISSIPPI begin with the letter M

Ex 7.3,10 - Chapter 7 Class 11 Permutations and Combinations (Term 2)

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Ex 7.3, 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

(a) Total letters = 11, I = 4, S = 4, P = 2. 

∴ The total number of permutations 7 

4s’s are together can be taken as 1 unit i.e.

∴ The number of permutations = \(\frac{8!}{4!\times2!}\) 

(b) 4s’s are not together = Total number of ways – 4s’s are together = \(\frac{11!}{4!\times4!\times2!}\) - \(\frac{8!}{4!\times2!}\)

(c) Begin with MISS: The remaining 7 letters can be arranged in \(\frac{7!}{3!\times(2!)^2}\)

(d) Begin with SIP: The remaining 8 letters (I = 3, S = 3) can be arranged in \(\frac{8!}{(3!)^2}\)

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Solution

Total letters of the word MISSISSIPPI = 11.Here M=1, I=4,S=4 and P=2∴ Number of permutations = 11!4! 4!2!= 11×10×9×8×7×6×5×4!4!×4×3×2×1×2×1= 34650When the four 'I's come together, then it becomes one letter so total number of letters in the word when all I"s come together = 8.∴ Number of permutations= 8!4!2!=8×7×6×5×4!4!×2×1=840Number of permutations when four I"s do not come together= 34650−840=33810.

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