**NCERT Solutions Class 11 Maths Chapter 7 Permutation And Combination** – Here are all the NCERT solutions for Class 11 Maths Chapter 7. This solution contains questions, answers, images, explanations of the complete chapter 7 titled Of Permutation And Combination taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Maths, then you must come across chapter 7 Permutation And Combination After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Maths Chapter 7 Permutation And Combination in one place.

## NCERT Solutions Class 11 Maths Chapter 7 Permutation And Combination

Here on **AglaSem Schools**, you can access to **NCERT Book Solutions** in free pdf for Maths for Class 11 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 7 Permutation And Combination , Maths, Class 11.

Class | 11 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 7 |

Chapter Name |
Permutation And Combination |

### NCERT Solutions Class 11 Maths chapter 7 Permutation And Combination

Class 11, Maths chapter 7, Permutation And Combination solutions are given below in PDF format. You can view them online or download PDF file for future use.

### Permutation And Combination

**NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Download**

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### Download NCERT Solutions Class 11 Maths chapter 7 Permutation And Combination In PDF Format

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### Question & Answer

Q.1:How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) repetition of the digits is allowed? (ii) repetition of the digits is not allowed?

Ans :(i)There will be as many ways as there are ways of filling 3 vacant places [] [] [] in succession by the given five digits. In this case, repetition of digits is allowed. Therefore, the units place can be filled in by any of the given five digits. Similarly, tens and hundreds digits can be filled in by any of the given five digits. Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed from the given digits is 5 × 5 × 5 = 125 (ii) In this case, repetition of digits is not allowed. Here, if units place is filled in first, then it can be filled by any of the given five digits. Therefore, the number of ways of filling the units place of the three-digit number is 5. Then, the tens place can be filled with any of the remaining four digits and the hundreds place can be filled with any of the remaining three digits. Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed without repeating the given digits is 5 × 4 × 3 = 60

Q.2:How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

Ans :There will be as many ways as there are ways of filling 3 vacant places [] [] [] in succession by the given six digits. In this case, the units place can be filled by 2 or 4 or 6 only i.e., the units place can be filled in 3 ways. The tens place can be filled by any of the 6 digits in 6 different ways and also the hundreds place can be filled by any of the 6 digits in 6 different ways, as the digits can be repeated. Therefore, by multiplication principle, the required number of three digit even numbers is 3 × 6 × 6 = 108

Q.3:How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

Ans :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.

Q.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?

Ans :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 by [6] [7] [] [] [] the digits 0 – 9, keeping in mind that the digits cannot be repeated. The units place can be filled by any of the digits from 0 – 9, except digits 6 and 7. 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

Q.5:A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Ans :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 2 × 2 × 2 = 8

## NCERT / CBSE Book for Class 11 Maths

You can download the NCERT Book for Class 11 Maths in PDF format for free. Otherwise you can also buy it easily online.

- Click here for NCERT Book for Class 11 Maths
- Click here to buy NCERT Book for Class 11 Maths

### All NCERT Solutions Class 11

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### All NCERT Solutions

You can also check out NCERT Solutions of other classes here. Click on the class number below to go to relevant NCERT Solutions of Class 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

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