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Looking for Class 11 Maths Exercise 7.1 solutions? You’re in the right place! This section provides clear and detailed solutions for all the questions from Exercise 7.1 of Chapter 7 – Permutations and Combinations. Aligned with the latest NCERT syllabus, these solutions introduce the fundamental principles of counting and help you understand the basics of permutations. Whether you’re referring to the Class 11 Ch 7 Exercise 7.1 solutions, reviewing NCERT Exemplar Class 11 Maths, or strengthening your grasp of Permutation and Combination Class 11, these step-by-step answers will guide you effectively. View or download the NCERT Solutions for Class 11 Maths Chapter 7 now and start building your confidence in solving problems with ease!
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Exercise 7.1
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?
(ii) repetition of the digits is not allowed?
Answer
(i) There will be as many ways as there are ways of filling 3 vacant in succession by the given five digits. In this case repetition of digit is allowed.
Therefore, the unit place can be filled in by any of the given five digits.
Similarity, tens and hundreds digits can be filled in by any of the given five digits.
Thus, by the multiplication principle, the numbers of ways in which three digit numbers can be formed from the given digit is ( 5 times 5 times 5=125 ).
(ii) in this case repetition of digits is not allowed. Here, if unit 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 unit 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.
1
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-digits even numbers can be formed from the digits (1 , 2,3,4,5,6 ) if the digits can be repeated?
Answer
There will be as many ways as there are ways of filling 3 vacant places in succession by the given 6 digits
In this case, the unit place can be filled by 2 or 4 or 6 only i.e., the unit 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 hundred 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 digits can be repeated.
Therefore, by multiplication principle, the required numbers of three digits even number is ( 3 times 6 times 6=108 )
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3. How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
Answer
There are as many codes as there 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 any 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 cab 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 times 9 times 8 times 7=5040 )
Hence, 5040 four letter codes can be formed using the first 10 letters of the English alphabet, 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?
Answer
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 ( underline{6}, underline{7}, underline{}, underline{ }) by the digit ( 0-9 ), keeping in mind that the digit cannot be repeated.
Therefore, the unit 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 hundred place can be filled in by any of the following 6 digits in 6 different ways.
Therefore, by multiplication principle, the required numbers 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?
Answer
When a coin is tossed once, the numbers 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 numbers of possible outcome are ( 2 times 2 times 2=8 )
6. Given 5 flags of different color, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
Answer
Each signal required the use of 2 flags.
There will be as many flags as there are always of filling in 2 vacant places in succession by the given 5 flags of different colors.
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 one of the remaining 4 different flags.
Thus, by multiplication principle the number of different signals that can be generated is ( 5 times 4=20 )
