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

## 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 the part of NCERT Solutions for Class 11 Maths. Here you can find the 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 places in succession by the five given digits.
(i) If 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 × 5 × 5, that is, 125.
(ii) If repetition is not allowed, then the first place can be filled in 5 different ways, the second place can be filled in 4 different ways and the third place can be filled in 3 different ways. Therefore, by the multiplication principle, the required number of 3-digit numbers is 5 × 4 × 3, that is, 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 in succession by the six given digits. In this case, we start filling in units place, because the options for this place are 2, 4 and 6 only and this can be done in 3 ways. The tens and the hundreds place can be filled in 6 different ways. Therefore, by the multiplication principle, the required number of 3-digit even numbers is 6 × 6 × 3, that is, 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 in succession by the first 10 letters of the English alphabet.

If repetition is not allowed, then the first place can be filled in 10 different ways, the second place can be filled in 9 different ways, the third place can be filled in 8 different ways and the fourth place can be filled in 7 different ways. Therefore, by the multiplication principle, the required number of 4-letter codes is 10 × 9 × 8 × 7, that is, 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, which start with 67, can be constructed using the digits 0 to 9.

If repetition is not allowed, then at the first and second places 6 and 7 are fixed, respectively. Therefore, the third, fourth and fifth places can be filled in 8, 7 and 6 ways, respectively. So, by the multiplication principle, the required number of 5-digit telephone numbers is 8 × 7 × 6, that is, 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, that is, head or tail. When the coin is tossed three times, then the total number of possible outcomes are 2 × 2 × 2, that is, 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 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 × 4, that is, 20.

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