     1. Chapter 7 Class 11 Permutations and Combinations
2. Serial order wise
3. Ex 7.3

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 𝑛!/𝑝1!𝑝2!𝑝3! Total number of letters = n = 10 & Since, 2T ⇒ 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! × 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 × 10!/2! Similarly, S can be before P Number of arrangements when S is before P = 7 × 10!/2! Total number of arrangements = 7 × 10!/2! + 7 × 10!/2! = 2 × 7 × 10!/2! = 2 × 7 × 10!/((2 × 1)) = 7 × 10! = 25401600

Ex 7.3 