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# MCQ - Permutations and Combinations CA Foundation Notes | EduRev

## CA Foundation : MCQ - Permutations and Combinations CA Foundation Notes | EduRev

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CPT Section D Quantitative Aptitude Chapter 5
Preethi Rathi

Page 2

CPT Section D Quantitative Aptitude Chapter 5
Preethi Rathi

Sol : The word DRAUGHT consists of 7 letters of which 5 are
consonants and two are vowels. Two Vowels should not be seperated.
2 vowels= 1 unit=     2! ? 2 ways-----(1)
remaning units=7-2+1=6 .the total number of ways of arranging them is
6p6 = 6! = 720 ways.----(2)
The total number of arrangements of word DRAUGHT
are(1)x(2)=2x720=1440 ways
Page 3

CPT Section D Quantitative Aptitude Chapter 5
Preethi Rathi

Sol : The word DRAUGHT consists of 7 letters of which 5 are
consonants and two are vowels. Two Vowels should not be seperated.
2 vowels= 1 unit=     2! ? 2 ways-----(1)
remaning units=7-2+1=6 .the total number of ways of arranging them is
6p6 = 6! = 720 ways.----(2)
The total number of arrangements of word DRAUGHT
are(1)x(2)=2x720=1440 ways
SOL: 6 economics books can be arranged among themselves
in 6! Ways---(1),3 mathematics books can be arranged in 3!
Ways –(2)and the 2 books on accountancy can be arranged
In 2! Ways—(3).
. Now there are three units. These 3 units can be arranged
in 3! Ways.----(4)
Total number of arrangements
(1)x(2)x(3)x(4)=3!x6!x3!x2!ways.
Page 4

CPT Section D Quantitative Aptitude Chapter 5
Preethi Rathi

Sol : The word DRAUGHT consists of 7 letters of which 5 are
consonants and two are vowels. Two Vowels should not be seperated.
2 vowels= 1 unit=     2! ? 2 ways-----(1)
remaning units=7-2+1=6 .the total number of ways of arranging them is
6p6 = 6! = 720 ways.----(2)
The total number of arrangements of word DRAUGHT
are(1)x(2)=2x720=1440 ways
SOL: 6 economics books can be arranged among themselves
in 6! Ways---(1),3 mathematics books can be arranged in 3!
Ways –(2)and the 2 books on accountancy can be arranged
In 2! Ways—(3).
. Now there are three units. These 3 units can be arranged
in 3! Ways.----(4)
Total number of arrangements
(1)x(2)x(3)x(4)=3!x6!x3!x2!ways.
SOL: (i) consider the 3 sisters as 1 unit =3!ways-----(1)                              .
total units=7-3+1= 5 units which can be arranged  in 5! Ways----(2) .  .
there fore total number of ways in which all the sisters sit
together=(1)x(2) =5!x3!=720 ways
(ii) 4brothers are aranged 4!ways-----(1)In this case,each sister must sit on
each side of the brothers. In between 4 brothers there  are 5  positions
as indicated below by upward arrow

B1          B2          B3        B4
3 sisters can sit in the 5 places in 5p3 ways.----(2)
Total number of ways =(1)x(2)
4!x5p3=60x24=1,440
Page 5

CPT Section D Quantitative Aptitude Chapter 5
Preethi Rathi

Sol : The word DRAUGHT consists of 7 letters of which 5 are
consonants and two are vowels. Two Vowels should not be seperated.
2 vowels= 1 unit=     2! ? 2 ways-----(1)
remaning units=7-2+1=6 .the total number of ways of arranging them is
6p6 = 6! = 720 ways.----(2)
The total number of arrangements of word DRAUGHT
are(1)x(2)=2x720=1440 ways
SOL: 6 economics books can be arranged among themselves
in 6! Ways---(1),3 mathematics books can be arranged in 3!
Ways –(2)and the 2 books on accountancy can be arranged
In 2! Ways—(3).
. Now there are three units. These 3 units can be arranged
in 3! Ways.----(4)
Total number of arrangements
(1)x(2)x(3)x(4)=3!x6!x3!x2!ways.
SOL: (i) consider the 3 sisters as 1 unit =3!ways-----(1)                              .
total units=7-3+1= 5 units which can be arranged  in 5! Ways----(2) .  .
there fore total number of ways in which all the sisters sit
together=(1)x(2) =5!x3!=720 ways
(ii) 4brothers are aranged 4!ways-----(1)In this case,each sister must sit on
each side of the brothers. In between 4 brothers there  are 5  positions
as indicated below by upward arrow

B1          B2          B3        B4
3 sisters can sit in the 5 places in 5p3 ways.----(2)
Total number of ways =(1)x(2)
4!x5p3=60x24=1,440
Sol: suppose that we have 11 chairs in a row and we want the 6 boys and
5 girls to be seated such that no two girls and no two boys are together. If
we number the chairs from left to right,the arrangements will be possible if
and only if boys occupy the odd places and girls occupy the even places
in the row.
The six odd places from 1 to 11 may filled in by 6 boys in 6p6 ways.---(1)
Similarly, the five even
Places from 2 to 10 may be filled in by 5 girls in 5p5 ways.---(2)
Hence, by the fundamental principle, the total number of required
arrangements
?(1)x(2)= 6p6 x 5p5 = 6! X 5! = 720x120 = 86400
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