Page 1 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=72+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=72+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=72+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=73+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=72+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=73+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 = 86400Read More
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