HOW TO SOLVE PROBLEMS ON PERMUTATIONS-LEARN SERIES
Hi Bankersdaily Aspirants,
IBPS RRB is going to be held shortly and you all know that even 1 mark is important for cracking the competitive exam.Aspirants,Aptitude Section is not that much tough as you think,There is trick to solve each problem in the Aptitude Section.And we are discussing each and every trick and shortcut for solving the Problems in each topic with its Possible Question from the Rudimentry level in which even Neophyte can crack the exam by looking at the article.
INTRODUCTION TO PERMUTATION:
Permutation is defined as the act of arranging all the members of a set into some sequence or order,if the set is already ordered,rearranging its elements,a process called permuting.This topic is also quite easy to solve and it is asked in 1 or 2marks in exam
In general P(n, r) means that the number of permutations of n things taken r at a time. We can either use reasoning to solve these types of permutation problems or we can use the permutation formula.
The formula for permutation is
Why we are moving for Permutation:
Lets consider an example:
1) 8 people(A,B,C,D,E,F,G,H) are participating in a Race and there were 3 Prizes Gold, Silver and Bronze .
2)A license plate begins with three letters. If the possible letters are A, B, C, D and E, how many different permutations of these letters can be made if no letter is used more than once?
TYPE 2:(INDISTINGUISHABLE ITEMS)
The number of different permutations of n objects where there are n1indistinguishable items, n2 indistinguishable items, … nk
Indistinguishable items, is
1)How many ways can the letters of the word MATHEMATICS be arranged?
In the Mathematics there are 11 letters in which M occur 2 times, T occur 2 times, A occur 2 times and the remaining word occur only once in the word Mathematics
Total number =11 and there are 3 Repeated Letters(M,T,A)
=11!/(2! *2!*2!) =11*10*9*8*7*6*5*4*3*2*1/(2*1)(2*1)(2*1) = 4989600 ways
TYPE 3 :
PERMUTATION WITH SOME RESTRICTIONS:
(a) Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is to be always included in each arrangement= r n-1 Pr-1
(b) Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is fixed: = n-1 Pr-1
(c) Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is never taken: = n-1 Pr.
(d) Number of permutations of ‘n’ things, taken ‘r’ at a time, when ‘m’ specified things always come together = m! x ( n-m+1) !
(e) Number of permutations of ‘n’ things, taken all at a time, when ‘m’ specified things always come together = n ! – [ m! x (n-m+1)! ]
Example: How many words can be formed with the letters of the word ‘OMEGA’ when:
(i) ‘O’ and ‘A’ occupying end places.
(ii) ‘E’ being always in the middle
(iii) Vowels occupying odd-places
(iv) Vowels being never together.
PERMUTATION WITH REPETITION AND WIHTOUT REPETITION:
1.If five digit 1,2,3,4,5 are being given and athree digit code has to be made from it if the repetition of digits is allowed then how many such codes can be formed?
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