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Class 11 RD Sharma Solutions – Chapter 16 Permutations – Exercise 16.5 | Set 1

Last Updated : 11 Feb, 2021
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Question 1. Find the number of words formed by permuting all the letters of the following words :

(i) INDEPENDENCE

(ii) INTERMEDIATE

(iii) ARRANGE

(iv) INDIA

(v) PAKISTAN

(vi) RUSSIA

(vii) SERIES

(viii) EXERCISES

(ix) CONSTANTINOPLE

Solution:

(i) Given: 12 letter word in which

N appear = 3 times

E appear = 4 times

D appear = 2 times

Remaining letters = once 

Number of permutations of these letters = 12! / (3! 4! 2! ) = 1663200

(ii) Given: Total 12 letters there 

I appear = twice

T appear =twice

E appear = thrice

Remaining = once

Number of permutations = 12! / (2! 2! 3!) = 19958400 

(iii) Given: 7 letters in which

A appear = 2 times 

R appear = 2 times

Remaining = once 

Number of permutations = 7! / (2! 2!) = 1260

(iv) Given: 5 letters in which

I appear = twice

Remaining = once

Number of permutations = 5! / 2! = 60 

(v) Given: 8 letters in which

A appear = 2 times

Remaining = once 

Number of permutations = 8! / 2! = 20160

(vi) Given: 6 letters in which 

S appear = twice 

Remaining = once 

Number of permutations = 6!/2! = 360

(vii) Given: 6 letters in which

S appear = twice

E appear = twice 

Remaining = once 

Number of permutations = 6! / (2! 2!) = 180

(viii) Given: 9 letters in which

E appear = 3 times 

S appear = 2 times

Remaining = once 

Number of permutations = 9! / (3! 2!) = 30240

(ix) Given: 14 letters in which

T appear = 2 times 

O appear = 2 times 

N appear = 3 times 

Remaining appear = 1 time 

Number of permutations = 14! / (2! 2! 3!) = 3632428800                          

Question 2. In how many ways can the letters of the word ‘ALGEBRA’ be arranged without changing the relative order of the vowels and consonants?

Solution:

Given: 3 vowels and 4 consonants present in the given word

Vowels (A, E, A) = 3!/2! = 3 ways of permutations

Consonants = 4! = 24 permutations    

Permutations so that the given condition is satisfied = 3 x 24 = 72 

Question 3. How many words can be formed with the letters of the word ‘UNIVERSITY,’ the vowels remaining together?

Solution:

4 vowels present can be arranged in = 4!/2! ways  

First vowel can have positions numbers 1 to 7 and 

remaining will follow after that in consecutive order = 7C1 = 7 ways 

Permutations of the consonants = 6! 

Total words that satisfy given conditions = 6! x 7 x 4!/2! = 60480    

Question 4. Find the total number of arrangements of the letters in the expression a3 b2 c4 when written at full length.

Solution:

In a3 b2 c4,

a appear = 3 times 

b appear = 2 times

c appear = 4 times 

So, the total arrangements = 9! / (3! 2! 4!) = 1260

Question 5. How many words can be formed with the letters of the word ‘PARALLEL’ so that all L’s do not come together?

Solution:

Total permutations = 8! / (3! 2!) = 3360

Number of words in which all L’s together (Replace 3 L by one symbol 

so this 1 symbol and 5 other letters left) 

= 6! / 2! = 360

Desired words = 3360 – 360 = 3000   

Question 6. How many words can be formed by arranging the letters of the word ‘MUMBAI’ so that all M’s come together?

Solution:

Total number of letters present in the MUMBAI word = 6

Replace 2 M’s by 1 symbol = 5 letters left there 

Number of words = 5! = 120 

Question 7. How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that the odd digits always occupy the odd places?

Solution:

Given: 4 digits are odd and 4 odd positions  

Arrange these 4 odd digits in 4 positions = 4!/(2! 2!) = 6     

Arrange remaining 3 digits = 3!/2! = 3 

Total permutations = 6 x 3 = 18

Question 8. How many different signals can be made from 4 red, 2 white, and 3 green flags by arranging all of them vertically on a flagstaff?

Solution:

Given: Total number of flags = 9 flags. In which 

red appear = 4

white appear = 2

green appear = 3 

So, the number of different signals = 9! / (4! 2! 3!) = 1260

Question 9. How many numbers of four digits can be formed with the digits 1, 3, 3, 0?

Solution:

Given: Total number = 4 digits in which

3 appear = twice 

Total different numbers = 4! / 2! = 12

But it will also have some 3-digit. 

3- digit numbers = 3!/2! = 3 

4-digit numbers = 12 – 3 = 9 

Question 10. In how many ways can the letters of the word ‘ARRANGE’ be arranged so that the two R’s are never together?

Solution:

Total permutations of the letters of given word = 7! / (2! 2!) = 1260   

Permutations with R together (replace two Rs by 1 letter) = 6! / 2! = 360 

Total permutations that satisfies given criteria = 1260 – 360 = 900

Question 11. How many different numbers, greater than 50000 can be formed with the digits 0, 1, 1, 5, 9.

Solution:

At first position, only 5 or 9 can be there = 2 ways 

Remaining 4 digits can be arranged at 4 positions in 4!/2! ways 

Number of different numbers = 2 x 4!/2! = 24    

Question 12. How many words can be formed from the letters of the word ‘SERIES’ which start with S and end with S?

Solution:

At first and last position, S and S are fixed so 1 way for these two S. 

At 2nd, 3rd, 4th, 5th positions = E, R, I, E 

so, number of possible arrangements = 4!/2! = 12

So, desired number of such words = 1 x 4!/2! = 12  

Question 13. How many permutations of the letters of the word ‘MADHUBANI’ do not begin with M but end with I?

Solution:

We fix an I at end so effectively first 8 letters are left to be dealt with

Total number of words that end with I = 8!/2!

Number of words that start with M letter = 7!/2!

Number of desired words = 8!/2! – 7!/2! = 17640      

Question 14. Find the number of numbers, greater than a million that can be formed with the digit 2, 3, 0, 3, 4, 2, 3?

Solution:

To have a number greater than a million it should be at least of 7 length. 

So no number starting with digit 0 can be counted. 

Total arrangements of given digits are = 7! / (3! 2!) = 420

Arrangements having 0 at first position = 6! / (3! 2!) = 60

Desired arrangements = 420 – 60 = 360 


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