By: R Chandrakala on November 13, 2017
Permutation and Combination is a topic which every student finds it pretty tough. Questions from this topic appear in an almost equal proportion in almost all of the competitive examinations. Every question in the Permutation and Combination section has always more than one way of solving the question. The point here is, however, to solve the question in a minimal time frame.
Here are a few tips to keep in mind before proceeding on to preparing this topic for CAT:
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The Candidate must be clear with all the basic concepts and formulae in this case.
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The Candidate must go through all related notes, relevant training material or the reference material for CAT Examination.
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The candidate must also try to solve questions with more than one method in place to see which of the method takes less time.
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Practice a lot of questions to gain a better understanding of this topic.
Here are a few examples of problems for the view of students:
Example: Enlist the number of four-digit numbers that can be formed from digits 1,2,3,4,5,6 (Repetition of Digits are not allowed)?
Answer: Thousands Place is usually filled with 6 methods. Likewise, Hundreds Place can be filled in 5 ways. Keeping this in mind, the answer would be calculated as 6*5*4*3 = 360.
Example: An individual has 6 friends who would be invited to dinner by way of invitation cards and he has 3 servants. In how many ways, can he extend the invitation card?
Answer: Friend 1 has 3 options to receive the invite, from Servant 1 or Servant 2 or Servant 3. Likewise for Friend 2 or Friend 3. The answer would be 3 raised to the power of 6 i.e. 729 ways.
Example: There are 10 questions in an examination. In how many ways will a person attempt at least one question?
Answer: The person can attempt from 1 to 10 questions in 10C1= 10 ways. 2 questions out of 10 questions in a similar way can be attempted in 10C2 ways = 45. On similar lines, ten questions can be attempted in 10C10 ways. The total number of ways would be 10+45+120...+1 =1023
Alternately, every question has 2 options irrespective of being attempted. For 10 questions with 2 options each, is 2 raised to the power of 10 equalling 1023 minus one case wherein the candidate would not be attempting the paper at all.
