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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 letters are repeating Hence we use this formula ๐!/๐1!๐2!๐3! Total number of letters = n = 10 & Since, 2T โด p1 = 2 Now, Total arrangements = 10!/2! = 1814400 Ex 7.3, 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 Letโs 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 โ 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 = ๐!/๐1!๐2!๐3! Total letters = n = 8 As 2T โด p1 = 2 Total arrangements = 8!/2! Hence, Total number of arrangement = 8!/2! ร 120 = 2419200 Ex 7.3, 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 ร 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 = ๐!/๐1!๐2!๐3! Number of letters = 10 n = 10 Here 2Tโs p1 = 2 Numer of arrangements = 10!/2! Thus, Total number of arrangements = 14 ร 10!/2! = 14 ร 10!/((2 ร 1)) = 7 ร 10! = 25401600
Solution: We know that the number of arrangements (permutations) that can be made out of n things out of which there are p, q, r, ... number of repetitions = n! / [p! q! r! ...]. In the given word PERMUTATIONS, No. of T's = 2 (i) It is given that words start with P and end with S. So the letters in the first and last positions are fixed. The middle 10 positions have to be filled with the remaining 10 letters among which there are 2 Ts (which are repeated). No. of words = 10!/2! = 18,14,400. Click here to see the formula behind it. (ii) vowels are all together, We know that there 5 vowels in the given letter. Then it becomes P, R, M, T, T, N, S, \(\fbox{EUAIO}\), so there are 8 units among which there are 2 Ts'. Also, vowels can be interchanged within themselves in 5! ways. Thus, the number of words = 8!/2!ร 5! = 24,19,200. (iii) It is given that there are always 4 letters between P and S. So P and S can take the following positions respectively.
There are in total 7 ways in which there are 4 letters between P and S. If we interchange P and S in the above table, we get 7 more ways of placing P and S. Thus, total number of ways in which P and S can be placed = 7 + 7 = 14. Now, the remaining 10 positions have to be filled with the remaining 10 letters among which there are 2 Ts. So No. of ways for filling the remaining 10 positions = 10!/2!. Total no. of ways = 14ร10!/2! = 2,54,01,600 NCERT Solutions Class 11 Maths Chapter 7 Exercise 7.3 Question 11 In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end with S, (ii) vowels are all together, (iii) there are always 4 letters between P and S?Summary: (i) No. of ways = 18,14,400; (ii) If we consider all the 5 vowels as 1 unit, then no. of ways = 24,19,200; (iii) Total no. of ways = 2,54,01,600 Open in App Suggest Corrections 14 |
In how many ways can the letters of PERMUTATIONS be arranged if the vowels are all together
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