NCERT Solutions for Class 11 Maths Chapter 6 Permutations And Combinations Ex 6.1

Exercise 6.1

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?

Ans: There will be however many ways as there are methods of filling 3 empty spots in progression by the given five digits. In this case, repetition of digits is allowed.

Therefore, the units spot can be filled in by any of the given five digits.

Similarity, tens and hundred digits can be filled in by any of the given five digits.

Hence, by the multiplication principle, the number of ways in which three-digit numbers can be formed from the given digits is $5 \times 5 \times 5 = 125$.

(ii) Repetition of the digits is not allowed?

Ans: In this case, repetition of digits is not allowed. Here, on the off chance that unit place is filled in first, it very well may be filled by any of the given five digits.

Therefore, the number of ways of filling the units spot of the three-digit number is 5.

Then, at that point, the tens spot can be filled with any of the leftover four digits and the hundreds spot can be filled with any of the excess 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 $5 \times 4 \times 3 = 60$.

2. How many 3-digit even numbers can be formed 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.

For this situation, the units spot can be filled by 2 or 4 or 6 just i.e., the units spot can be filled in 3 ways.

The tens spot can be filled by any of the 6 digits in 6 distinctive manners and furthermore the hundreds spot can be filled by any of the 6 digits in 6 diverse manners, as the digits can be repeated. Thus, by multiplication principle, the required number of three digit even numbers is $3 \times 6 \times 6 = 108$.

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 primary spot can be filled in 10 distinctive manners by any of the initial 10 letters of the English letters in order following which, the runner up can be filled in by any of the leftover letters in 9 diverse manners. The third spot can be filled in by any of the excess 8 letters in 8 diverse manners and the fourth spot can be filled in by any of the leftover 7 letters in 7 distinctive manners.

Therefore, by multiplication principle, the required numbers of ways in which 4 empty places can be filled is $10 \times 9 \times 8 \times 7 = 5040$.

Hence, 5040 four-letter codes can be shaped utilizing the initial 10 letters of the English letters in order, if no letter is repeated.

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: Since, it is given that the 5-digit telephone numbers always start with 67.

Therefore, there will be as many phone numbers as there are methods of filling 3 empty spots 6, 7, __, __, __ by the digits 0 – 9, remembering that the digits cannot be repeated.

In this way, the units spot can be filled in 8 diverse manners following which, the tens spot can be filled in by any of the leftover 7 digits in 7 distinctive manners, and the hundreds spot can be filled in by any of the excess 6 digits in 6 distinctive manners. Therefore, by multiplication principle, the required number of ways in which 5-digit telephone numbers can be constructed is $8 \times 7 \times 6 = 336$.

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 $2 \times 2 \times 2 = 8$.

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?

Ans: Since, each signal requires the use of 2 flags.

There will be however many flags as there are methods of filling in 2 empty spots in progression by the given 5 flags of various colours. 

The upper empty spot can be filled in 5 distinctive manners by any of the 5 flags following which, the lower empty spot can be filled in 4 diverse manners by any of the leftover 4 unique flags.

Thus, by multiplication principle, the number of different signals that can be generated is $5 \times 4 = 20$.

NCERT Solution Class 11 Maths of Chapter 1 All Exercises

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Exercise 6.1

Opting for the NCERT solutions for Ex 6.1 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 6.1 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 6 Exercise 6.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 11 Maths Chapter 6 Exercise 6.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

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