Ex 6.3

Chapter 6 Class 11 Permutations and Combinations
Serial order wise

Transcript

Ex 6.3, 10 In how many of the distinct permutations of the letters in MISSISSIPPI do the four Iās not come together? Total number of permutation of 4I not coming together = Total permutation ā 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 = š!/š1!š2!š3! 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 = š!/š1!š2!š3! = 11!/(4! 4! 2!) = (11 Ć 10 Ć 9 Ć 8 Ć 7 Ć 6 Ć 5 Ć 4!)/((4 Ć 3 Ć 2 Ć 1) (4!)Ć(2 Ć 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 = š!/š1!š2! Number of letters = 8 ā“ n = 8 Since there are 4 S & 2 P p1 = 4, p2 = 2, Number of permutation with 4I together = š!/š1!š2! = 8!/(4! 2!) = 840 Now, Total number of permutation of 4I not coming together = Total permutation ā total permutation of I coming together = 34650 ā 840 = 33810