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Class XI Chapter 7 –

Class XI Chapter 7 – Permutation and Combination Maths ______________________________________________________________________________ Solution 9: There are 6 different letters in the word MONDAY. (i) Number of 4-letter words that can be formed from the letters of the word MONDAY, without repetition of letters, is the number of permutations of 6 different objects taken 4 at a time, which is 6 P 4 . Thus, required number of words that can be formed using 4 letters at a time is 6 6! 6! 65432! P4 6543 360 6 4 ! 2! 2! (ii) Number of words that can be formed by using all the letters of the word MONDAY at a time is the number of permutations of 6 different objects taken 6 at a time, which is 6 P6 6! Thus, required number of words that can be formed when all letters are used at a time 6! 654321 720 (iii) In the given word, there are 2 different vowels, which have to occupy the rightmost place of the words formed. This can be done only in 2 ways. Since the letters cannot be repeated and the rightmost place is already occupied with a letter (which is a vowel), the remaining five places are to be filled by the remaining 5 letters. This can be done in 5! ways. Thus, in this case, required number of words that can be formed is 5! 2 120 2 240 . Question 10: In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together? Solution 10: In the given word MISSISSIPPI, I appears 4 times, S appears 4 times, P appears 2 times, and M appears just once. Therefore, number of distinct permutations of the letters in the given word 11! 4!4!2! 111098 7 65 4! 4! 43 21 21 111098765 43 21 21 34650 There are 4 Is in the given word. When they occur together, they are treated as single object IIII for the time being. This single object together with the remaining 7 objects will account Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.

Class XI Chapter 7 – Permutation and Combination Maths ______________________________________________________________________________ for 8 objects. These 8 objects in which there are 4 Ss and 2 Ps can be arranged in 8! 4!2! ways i.e., 840 ways. Number of arrangements where all Is occur together = 840 Thus, number of distinct permutations of the letters in MISSISSIPPI in which four Is do not come together . 34650 840 33810 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? Solution 11: In the word PERMUTATIONS, there are 2 Ts and all the other letters appear only once. (i) If P and S are fixed at the extreme ends (P at the left end and S at the right end), then 10 letters are left. Hence, in this case, required number of arrangements 10! 1814400 2! (ii) There are 5 vowels in the given word, each appearing only once. Since they have to always occur together, they are treated as a single object for the time being. This single object together with the remaining 7 objects will account for 8 objects. These 8 objects in which there are 2 Ts can be arranged in 8! 2! ways. Corresponding to each of these arrangements, the 5 different vowels can be arranged in 5! ways. Therefore, by multiplication principle, required number of arrangements in this case 8! 5! 2419200 2! (iii) The letters have to be arranged in such a way that there are always 4 letters between P and S. Therefore, in a way, the places of P and S are fixed. The remaining 10 letters in which there are 2 Ts can be arranged in 10! 2! ways. Also, the letters P and S can be placed such that there are 4 letters between them in 2 7 14 ways. Therefore, by multiplication principle, required number of arrangements in this case 10! 14 25401600 2! . Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.

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    Class XI Chapter 4 - Principle of M

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    Class XI Chapter 11 - Conic Section

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    Class XI Chapter 12 - Introduction

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    Class XI Chapter 13 - Limits and De

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    Class XI Chapter 15 - Statistics Ma

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