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

• Last Updated : 11 Feb, 2021

### 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 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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