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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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FAQs on MCQ - Permutations and Combinations - Quantitative Aptitude for CA Foundation

1. What is the difference between permutations and combinations?
Ans. Permutations and combinations are both concepts used in mathematics to count the number of possible arrangements or selections. The main difference between them lies in the order of selection. In permutations, the order of the elements matters. For example, if we have three different objects A, B, and C, the permutations of selecting two objects would be AB, AC, BA, BC, CA, and CB. On the other hand, in combinations, the order does not matter. Using the same example, the combinations of selecting two objects would be AB, AC, and BC.
2. How do we calculate permutations?
Ans. The formula to calculate permutations is given by nPr = n! / (n - r)!, where n represents the total number of objects and r represents the number of objects being selected. For example, if we have 5 objects and we want to select 3 of them in a specific order, we would calculate it as 5P3 = 5! / (5 - 3)! = 5! / 2! = 60. Hence, there are 60 different permutations possible in this scenario.
3. How do we calculate combinations?
Ans. The formula to calculate combinations is given by nCr = n! / (r! * (n - r)!), where n represents the total number of objects and r represents the number of objects being selected. For example, if we have 5 objects and we want to select 3 of them without considering the order, we would calculate it as 5C3 = 5! / (3! * (5 - 3)!) = 5! / (3! * 2!) = 10. Hence, there are 10 different combinations possible in this scenario.
4. How are permutations and combinations used in real-life scenarios?
Ans. Permutations and combinations are used in various real-life scenarios. Some examples include: - Lottery systems, where the order of the numbers drawn matters (permutations). - Planning a menu with different courses and options (combinations). - Creating passwords with different characters and lengths (permutations). - Arranging a playlist of songs in a specific order (permutations). - Selecting a committee from a group of people without considering their positions (combinations). These concepts help in systematically counting the possibilities and making informed decisions.
5. Are there any shortcuts or tricks to solve permutations and combinations problems quickly?
Ans. Yes, there are certain shortcuts and tricks to solve permutations and combinations problems quickly. Some of them include: - Using factorial notation to simplify calculations. - Identifying patterns and symmetry in the problem. - Utilizing the concept of "n choose r" to solve combinations. - Understanding the concept of "n minus r" to solve permutations. Practicing various problems and understanding the underlying concepts will enhance problem-solving skills and make the process more efficient.
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