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# CLASS_11_MATHS_SOLUTIONS_NCERT

## Class XI Chapter 7 –

Class XI Chapter 7 – Permutation and Combination Maths ______________________________________________________________________________ Question 3: How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated? Solution 3: There are as many codes as there are ways of filling 4 vacant places in succession by the first 10 letters of the English alphabet, keeping in mind that the repetition of letters is not allowed. The first place can be filled in 10 different ways by any of the first 10 letters of the English alphabet following which, the second place can be filled in by any of the remaining letters in 9 different ways. The third place can be filled in by any of the remaining 8 letters in 8 different ways and the fourth place can be filled in by any of the remaining 7 letters in 7 different ways. Therefore, by multiplication principle, the required numbers of ways in which 4 vacant places can be filled is 10 9 8 7 5040 Hence, 5040 four-letter codes can be formed using the first 10 letters of the English alphabet, if no letter is repeated. Question 4: How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once? Solution 4: It is given that the 5-digit telephone numbers always start with 67. Therefore, there will be as many phone numbers as there are ways of filling 3 vacant places 6, 7, __, __, __ by the digits 0 – 9, keeping in mind that the digits cannot be repeated. Therefore, the units place can be filled in 8 different ways following which, the tens place can be filled in by any of the remaining 7 digits in 7 different ways, and the hundreds place can be filled in by any of the remaining 6 digits in 6 different ways. Therefore, by multiplication principle, the required number of ways in which 5-digit telephone numbers can be constructed is 8 7 6 336 Question 5: A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there? Solution 5: When a coin is tossed once, the number of outcomes is 2 (Head and tail) i.e., in each throw, the number of ways of showing a different face is 2. Thus, by multiplication principle, the required number of possible outcomes is 222 8 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 ______________________________________________________________________________ Question 6: Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other? Solution 6: Each signal requires the use of 2 flags. There will be as many flags as there are ways of filling in 2 vacant places in succession by the given 5 flags of different colours. The upper vacant place can be filled in 5 different ways by any one of the 5 flags following which, the lower vacant place can be filled in 4 different ways by any one of the remaining 4 different flags. Thus, by multiplication principle, the number of different signals that can be generated is 54 20 . Exercise 7.2 Question 1: Evaluate (i) 8! (ii) 4! – 3! Solution 1: (i) 8! 1 2 3 4 5 6 7 8 40320 (ii) 4! 1 2 3 4 24 3! 123 6 4! 3! 24 6 18 Question 2: Is 4! 3! 7!? Solution 2: 3! 123 6 4! 1234 24 3! 4! 6 24 307! 1 23 4567 5040 3! 4! 7! 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 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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