Ex 6.3,11 - Chapter 6 Class 11 Permutations and Combinations

Transcript

Ex 6.3, 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 letters are repeating Hence we use this formula 𝑛!/𝑝1!𝑝2!𝑝3! Total number of letters = n = 10 & Since, 2T ∴ p1 = 2 Now, Total arrangements = 10!/2! = 1814400 Ex 6.3, 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) We treat as a single object So our letters become Let’s arrange them now Arranging 5 vowels Since vowels are coming together, they can be and so on Total letter in AEIOU = 5 Total permutations of 5 letters = 5P5 = 5!/(5 − 5)! = 5!/0! = 5!/1 = 120 Arranging remaining letters Numbers we need to arrange = 7 + 1 = 8 Here are 2T Since letter are repeating, We use this formula = 𝑛!/𝑝1!𝑝2!𝑝3! Total letters = n = 8 As 2T ∴ p1 = 2 Total arrangements = 8!/2! Hence, Total number of arrangement = 8!/2! × 120 = 2419200 Ex 6.3, 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? Not possible as there are only 12 digits So, there are total 7 cases where there are 4 digits between P and S But that is when P is before S There can also be a case where S is before P Thus, Total cases = 7 × 2 = 14 cases Now, lets find Permutation of letters in 1 case Since Position of P & S are fixed We need to arrange remaining letters i.e. (E, R, M, U, T, A, T, I, O, N) Since, T is repeating, we use this formula = 𝑛!/𝑝1!𝑝2!𝑝3! Number of letters = 10 n = 10 Here 2T’s p1 = 2 Numer of arrangements = 10!/2! Thus, Total number of arrangements = 14 × 10!/2! = 14 × 10!/((2 × 1)) = 7 × 10! = 25401600