# RD SHARMA Solutions for Class 11-science Maths Chapter 16 - Permutations

## Chapter 16 - Permutations Exercise Ex. 16.1

## Chapter 16 - Permutations Exercise Ex. 16.2

How many three-digit numbers are there whit no digit repeated?

**One-digit odd number:**

3 possible ways are there. These numbers are 3 or 5 or 7.

**Two-digit odd number:**

Tens place can be filled up by 3 ways (using any of the digit among 3, 5 and 7) and then the ones place can be filled in any of the remaining 2 digits.

So, there are 3 2 = 6 such 2-digit numbers.

**Three-digit odd number:**

Ignore the presence of zero at ones place for some instance.

Hundreds place can be filled up in 3 ways (using any of any of the digit among 3, 5 and 7), then tens place in 3 ways by using remaining 3 digits (after using a digit, there will be three digits) and then the ones place in 2 ways.

So, there are a total of 3 3 2 = 18 numbers of 3-digit numbers which includes both odd and even numbers (ones place digit are zero). In order to get the odd numbers, it is required to ignore the even numbers i.e. numbers ending with zero.

To obtain the even 3-digit numbers, ones place can be filled up in 1 way (only 0 to be filled), hundreds place in 3 ways (using any of the digit among 3, 5, 7) and then tens place in 2 ways (using remaining 2 digits after filling up hundreds place).

So, there are a total of 1 3 2 = 6 even 3-digit numbers using the digits 0, 3, 5 and 7 (repetition not allowed)

So, number of three-digit odd numbers using the digits 0, 3, 5 and 7 (repetition not allowed) = 18 - 6 = 12.

Therefore, odd numbers less than 1000 can be formed by using the digits 0, 3, 5, 7 when repetition of digits is not allowed are 3 + 6 + 12 = 21.

Serial numbers for an item produced in a factory are to be made using two letters followed by four digits (0 to 9). If the letters are to be are to be taken from six letters of English alphabet without repetition and the digits are also not repeated in a serial number, how many serial numbers are possible?

How many for digit natural numbers not exceeding 4321 can be formed with the digits 1, 2, 3, and 4, if the digits can repeat?

The given digits are 1, 2, 3 and 4. These digits can be repeated while forming the numbers. So, number of required four digit natural numbers can be found as follows.

Consider four digit natural numbers whose digit at thousandths place is 1.

Here, hundredths place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)

Similarly, tens place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)

Ones place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)

Number of four digit natural numbers whose digit at thousandths place is 1 = 4 4 4 = 64

Similarly, number of four digit natural numbers whose digit at thousandths place is 2 = 4 4 4 = 64

Now, consider four digit natural numbers whose digit at thousandths place is 4:

Here, if the digit at hundredths place is 1, then tens place can be filled in 4 ways and ones place can also be filled in 4 ways.

If the digit at hundredths place is 2, then tens place can be filled in 4 ways and ones place can also be filled in 4 ways.

If the digit at hundredths place is 3 and the digit at tens place is 1, then ones place can be filled in 4 ways.

If the digit at hundredths place is 3 and the digit at tens place is 2, then ones place can be filled only in 1 way so that the number formed is not exceeding 4321.

Number of four digit natural numbers not exceeding 4321 and digit at thousandths place is 3 = 4 4 + 4 4 + 4 + 1 = 37

Thus, required number of four digit natural numbers not exceeding 4321 is 64 + 64 + 64 + 37 = 229.

How many numbers of six digits can be formed from the digits 0, 1, 3, 5, 7, and 9 when on digit is repeated? How many of them are divisible by 10?

In how many ways can 5 different balls be distributed among three boxes?

There are 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated.

Each lamps has two possibilities either it can be switched on or off.

There are 10 lamps in the hall.

So the total numbers of possibilities are 2^{10}.

To illuminate the hall we require at least one lamp is to be switched on.

There is one possibility when all the lamps are switched off. If all the bulbs are switched off then hall will not be illuminated.

So the number of ways in which the hall can be illuminated is 2^{10}-1.

## Chapter 16 - Permutations Exercise Ex. 16.3

All the letters of the word ‘EAMCOT’ are arranged in different possible ways. Find the number of arrangements in which no two vowels are adjacent to each other.

## Chapter 16 - Permutations Exercise Ex. 16.4

## Chapter 16 - Permutations Exercise Ex. 16.5

How many permulations of the letters of the world 'MADHUBANI' do not begain with M but end with I?

## Chapter 16 - Permutations Exercise Ex. 16VSAQ

First we will find the least factorial term divisible by 14.

As 7!=7x6x5! is divisible by 14 leaving remainder zero.

Hence terms 7! onwards can be written as multiple of 7!.

8!=8x7!, 9!=9x8x7!... like ways 200! can also be written as multiple of 7!.

So all the terms 7! onwards are divisible by 14 leaving remainder zero.

1! + 2! + 3! + 4! + 5! + 6!

=1+2+6+24+120+720

=873

Hence remainder obtained when 1! + 2! + 3! + ….+ 200! is divided by 14 is 5.

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