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# RD Chapter 16- Permutations Ex-16.5 Interview Questions Answers

### Related Subjects

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

(i) INDEPENDENCE

There are 12 lettersin the word ‘INDEPENDENCE’ out of which 2 are D’s, 3 are N’s, 4 are E’s and therest all are distinct.

So by using theformula,

n!/ (p! × q! × r!)

total number ofarrangements = 12! / (2! 3! 4!)

=[12×11×10×9×8×7×6×5×4×3×2×1] / (2! 3! 4!)

= [12×11×10×9×8×7×6×5]/ (2×1×3×2×1)

= 11×10×9×8×7×6×5

= 1663200

(ii) INTERMEDIATE

There are 12 lettersin the word ‘INTERMEDIATE’ out of which 2 are I’s, 2 are T’s, 3 are E’s and therest all are distinct.

So by using theformula,

n!/ (p! × q! × r!)

total number ofarrangements = 12! / (2! 2! 3!)

= [12×11×10×9×8×7×6×5×4×3×2×1]/ (2! 2! 3!)

=[12×11×10×9×8×7×6×5×3×2×1] / (3!)

= 12×11×10×9×8×7×6×5

= 19958400

(iii) ARRANGE

There are 7 letters inthe word ‘ARRANGE’ out of which 2 are A’s, 2 are R’s and the rest all aredistinct.

So by using theformula,

n!/ (p! × q! × r!)

total number ofarrangements = 7! / (2! 2!)

= [7×6×5×4×3×2×1] /(2! 2!)

= 7×6×5×3×2×1

= 1260

(iv) INDIA

There are 5 letters inthe word ‘INDIA’ out of which 2 are I’s and the rest all are distinct.

So by using theformula,

n!/ (p! × q! × r!)

total number ofarrangements = 5! / (2!)

= [5×4×3×2×1] / 2!

= 5×4×3

= 60

(v) PAKISTAN

There are 8 letters inthe word ‘PAKISTAN’ out of which 2 are A’s and the rest all are distinct.

So by using theformula,

n!/ (p! × q! × r!)

total number ofarrangements = 8! / (2!)

= [8×7×6×5×4×3×2×1] /2!

= 8×7×6×5×4×3

= 20160

(vi) RUSSIA

There are 6 letters inthe word ‘RUSSIA’ out of which 2 are S’s and the rest all are distinct.

So by using theformula,

n!/ (p! × q! × r!)

total number ofarrangements = 6! / (2!)

= [6×5×4×3×2×1] / 2!

= 6×5×4×3

= 360

(vii) SERIES

There are 6 letters inthe word ‘SERIES’ out of which 2 are S’s, 2 are E’s and the rest all aredistinct.

So by using theformula,

n!/ (p! × q! × r!)

total number ofarrangements = 6! / (2! 2!)

= [6×5×4×3×2×1] / (2!2!)

= 6×5×3×2×1

= 180

(viii) EXERCISES

There are 9 letters inthe word ‘EXERCISES’ out of which 3 are E’s, 2 are S’s and the rest all aredistinct.

So by using theformula,

n!/ (p! × q! × r!)

total number ofarrangements = 9! / (3! 2!)

= [9×8×7×6×5×4×3×2×1]/ (3! 2!)

= [9×8×7×6×5×4×3×2×1]/ (3×2×1×2×1)

= 9×8×7×5×4×3×1

= 30240

(ix) CONSTANTINOPLE

There are 14 lettersin the word ‘CONSTANTINOPLE’ out of which 2 are O’s, 3 are N’s, 2 are T’s andthe rest all are distinct.

So by using theformula,

n!/ (p! × q! × r!)

total number ofarrangements = 14! / (2! 3! 2!)

= 14!/ (2×1×3×2×1×2×1)

= 14! / 24

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?

Given:

The word ‘ALGEBRA’

There are 4 consonantsin the word ‘ALGEBRA’

The number of ways toarrange these consonants is 4P4 = 4!

There are 3 vowels inthe word ‘ALGEBRA’ of which, 2 are A’s

So vowels can bearranged in n!/ (p! × q! × r!) = 3! / 2! Ways

Hence, the requirednumber of arrangements = 4! × (3! / 2!)

= [4×3×2×1×3×2×1] /(2×1)

= 4×3×2×1×3 = 72

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

Given:

The word ‘UNIVERSITY’

There are 10 lettersin the word ‘UNIVERSITY’ out of which 2 are I’s

There are 4 vowels inthe word ‘UNIVERSITY’ out of which 2 are I’s

So these vowels can beput together in n!/ (p! × q! × r!) = 4! / 2! Ways

Let us consider these4 vowels as one letter, remaining 7 letters can be arranged in 7! Ways.

Hence, the requirednumber of arrangements = (4! / 2!) × 7!

=(4×3×2×1×7×6×5×4×3×2×1) / (2×1)

= 4×3×2×1×7×6×5×4×3

= 60480

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

There are 9 (i.epowers 3 + 2 + 4 = 9) objects in the expression a3 b2 c4 and there are 3 a’s, 2 b’s, 4c’s

So by using theformula,

n!/ (p! × q! × r!)

total number ofarrangements = 9! / (3! 2! 4!)

= [9×8×7×6×5×4×3×2×1]/ (3×2×1×2×1×4×3×2×1)

= 7×6×5×3×2×1

= 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?

Given:

The word ‘PARALLEL’

There are 8 letters inthe word ‘PARALLEL’ out of which 2 are A’s, 3 are L’s and the rest all aredistinct.

So by using theformula,

n!/ (p! × q! × r!)

total number ofarrangements = 8! / (2! 3!)

= [8×7×6×5×4×3×2×1] /(2×1×3×2×1)

= 8×7×5×4×3×1

= 3360

Now, let us considerall L’s together as one letter, so we have 6 letters out of which A repeats 2times and others are distinct.

These 6 letters can bearranged in 6! / 2! Ways.

The number of words inwhich all L’s come together = 6! / 2!

= [6×5×4×3×2×1] / (2×1)

= 6×5×4×3

= 360

So, now the number ofwords in which all L’s do not come together = total number of arrangements –The number of words in which all L’s come together

= 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?

Given:

The word ‘MUMBAI’

There are 6 letters inthe word ‘MUMBAI’ out of which 2 are M’s and the rest all are distinct.

So let us considerboth M’s together as one letter, the remaining 5 letters can be arranged in 5!Ways.

Total number ofarrangements = 5!

= 5×4×3×2×1

= 120

Hence, a total numberof words formed during the arrangement of letters of word MUMBAI such that allM’s remains together equals to 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?

Given:

The digits 1, 2, 3, 4,3, 2, 1

The total number ofdigits are 7.

There are 4 odd digits1,1,3,3 and 4 odd places (1,3,5,7)

So, the odd digits canbe arranged in odd places in n!/ (p! × q! × r!) = 4!/(2! 2!) ways.

The remaining evendigits 2,2,4 can be arranged in 3 even places in n!/ (p! × q! × r!) = 3!/2!Ways.

Hence, the totalnumber of digits = 4!/(2! 2!) × 3!/2!

= [4×3×2×1×3×2×1] /(2! 2! 2!)

= 3×2×1×3×1

= 18

Hence, the number ofways of arranging the digits such odd digits always occupies odd places isequals to 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?

Given:

Number of red flags =4

Number of white flags= 2

Number of green flags= 3

So there are total 9flags, out of which 4 are red, 2 are white, 3 are green

By using the formula,

n!/ (p! × q! × r!) =9! / (4! 2! 3!)

= [9×8×7×6×5×4!] /(4!×2×1×3×2×1)

= [9×8×7×6×5] /(2×3×2)

= 9×4×7×5

= 1260

Hence, 1260 differentsignals can be made.

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

Given:

The digits 1, 3, 3, 0

Total number of digits= 4

Digits of the sametype = 2

Total number of 4digit numbers = 4! / 2!

Where, zero cannot bethe first digit of the four digit numbers.

So, Total number of 3digit numbers = 3! / 2!

Total number ofNumbers = (4! / 2!) – (3! / 2!)

= [(4×3×2)/2] –[(3×2)/2]

= [4×3] – [3]

= 12 – 3

= 9

Hence, total number offour digit can be formed is 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?

There are 7 letters inthe word ‘ARRANGE’ out of which 2 are A’s, 2 are R’s and the rest all aredistinct.

So by using theformula,

n!/ (p! × q! × r!)

total number ofarrangements = 7! / (2! 2!)

= [7×6×5×4×3×2×1] /(2! 2!)

= 7×6×5×3×2×1

= 1260

Let us consider allR’s together as one letter, there are 6 letters remaining. Out of which 2 timesA repeats and others are distinct.

So these 6 letters canbe arranged in n!/ (p! × q! × r!) = 6!/2! Ways.

The number of words inwhich all R’s come together = 6! / 2!

= [6×5×4×3×2!] / 2!

= 6×5×4×3

= 360

So, now the number ofwords in which all L’s do not come together = total number of arrangements –The number of words in which all L’s come together

= 1260 – 360

= 900

Hence, the totalnumber of arrangements of word ARRANGE in such a way that not all R’s cometogether is 900.

krishan