Last updated at Jan. 10, 2019 by Teachoo

Transcript

Ex 7.3, 11 In how many ways can the letters of the word PERMUTATIONS be arranged if the words start with P and end with S Let first position be P & last position be S (both are fixed) Since letter are repeating Hence we use this formula ๐!/๐1!๐2!๐3! Total number of letters = n = 10 & Since, 2T โ p1 = 2 Total arrangements = 10!/2! = 181440 Ex7.3, 11 In how many ways can the letters of the word PERMUTATIONS be arranged if the (ii) vowels are all together, Vowels are a, e, i, o, u Vowels in word PERMUTATION = (E U A I O) All vowels comes together treat as a single object So our letters become We arrange them now Total number of arrangement = 8!/2! ร 120 = 2419200 Ex7.3, 11 In how many ways can the letters of the word PERMUTATIONS be arranged if the (iii) there are always 4 letters between P and S? Hence, Number of arrangements when P is before S = 7 ร 10!/2! Similarly, S can be before P Number of arrangements when S is before P = 7 ร 10!/2! Total number of arrangements = 7 ร 10!/2! + 7 ร 10!/2! = 2 ร 7 ร 10!/2! = 2 ร 7 ร 10!/((2 ร 1)) = 7 ร 10! = 25401600

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.