RD Chapter 16 Permutations Ex 16.5 Solutions
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
Answer - 1 : -
(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?
Answer - 2 : -
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?
Answer - 3 : -
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.
Answer - 4 : -
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?
Answer - 5 : -
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?
Answer - 6 : -
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?
Answer - 7 : -
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?
Answer - 8 : -
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?
Answer - 9 : -
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?
Answer - 10 : -
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.