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NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.1

NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.1 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.1.

Board CBSE
Textbook NCERT
Class Class 11
Subject Maths
Chapter Chapter 7
Chapter Name Permutations and Combinations
Exercise Ex 7.1
Number of Questions Solved 6
Category NCERT Solutions

NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.1

Ex 7.1 Class 11 Maths Question 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?
Solution.
There will be as many ways as there are ways of filling 3 vacant \boxed { } \boxed { } \boxed { } places in succession by the five given digits.
(i) When repetition is allowed then each place can be filled in five different ways. Therefore, by the multiplication principle the required number of 3- digit numbers is 5 x 5 x 5 i.e., 125.
(ii) When repetition is not allowed then first place can be filled in 5 different ways, second place can be filled in 4 different ways & third place can be filled in 3 different ways. Therefore by the multiplication principle the required number of three digit numbers is 5 x 4 x 3 i.e, 60.

Ex 7.1 Class 11 Maths Question 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?
Solution.
There will be as many ways as there are ways of filling 3 vacant places \boxed { } \boxed { } \boxed { } in succession by the 6 given digits. In this case we start filling in unit’s place, because the options for this place are 2, 4 & 6 only and this can be done in 3 ways. Ten’s and hundred’s place can be filled in 6 different ways. Therefore, by the multiplication principle, the required number of 3-digit even numbers is 6 x 6 x 3 i.e., 108.

Ex 7.1 Class 11 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.
There will be as many ways as there are ways of filling 4 vacant places \boxed { } \boxed { } \boxed { } \boxed { } in succession by the first 10 letters of the English alphabet, when repetition is not allowed then first place can be filled in 10 different ways, second place can be filled in 9 different ways, third place can be filled in 8 different ways and fourth place can be filled in 7 different ways. Therefore, by the multiplication principle the required number of 4 letter codes are 10 x 9 x 8 x 7 i.e., 5040.

Ex 7.1 Class 11 Maths 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.
The 5 digit telephone numbers of the form \boxed { 6 } \boxed { 7 } \boxed { } \boxed { } \boxed { } can be constructed using the digits 0 to 9. When repetition is not allowed then at first & second place 6 & 7 are fixed respectively. Therefore, third, fourth and fifth place can be filled in 8, 7 and 6 ways respectively. So, by the multiplication principle the required numbers of 5-digit telephone numbers is 8 x 7 x 6 i.e., 336.

Ex 7.1 Class 11 Maths Question 5.
A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
Solution.
When a coin is tossed there are two possible outcomes i.e. head or tail. When the coin is tossed three times then the total possible outcomes are 2 x 2 x 2 i.e., 8.

Ex 7.1 Class 11 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.
There will be as many signals as there are ways of filling in 2 vacant places \boxed { } \boxed { } in succession by the 5 flags of different colours. The upper vacant place can be filled in 5 different ways by anyone of the 5 flags ; following which the lower vacant place can be filled in 4 different ways by anyone of the remaining 4 different flags. Hence by the multiplication principle the required number of signals is 5 x 4 = 20.

 

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