PERMUTATION

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Saumyojit

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CASE 1:KELLLY same as KEL1L2L3Y All l's are same; all the arangemnts of one particular form (KELLLY)
1:KEL1L2L3Y(Original set) 2:KEL2L1L3Y 3:KEL2L3L1Y 4:KEL1L3L2Y 5:KEL3L2L1Y 6:KEL3L1L2Y -->IF we change the position of L we will get 5 duplicates for each 1 distinct arrangement.
Total arrangements of KELLLY with repetions=npr=6factorial
6 factrial consists of each unique arrangements upto total n unqiue sets{x1,x2...Xn}+5 duplicates of every each unique arrangements 5{x1,x2...Xn}={x1,x2...Xn}+5{x1,x2...Xn}

=>{x1,x2...Xn}+5{x1,x2...Xn}=6 factorial
=>6{x1,x2...Xn}=6 factorial
=>x1...Xn=6 factorail/6 =120 arrangements
n=120 n denotes total no of disitnct permutations
Is my method of explaining this thing to myself is right?
CASE 2:If i try the same method with this word
REERR = R1E1E2R2R3

1:R2E1E2R1R3 2:R2E1E2R3R1 3:R3E1E2R1R2 4:R3E1E2R2R1 5:R1E1E2R3R2
only changing r position will give me extra 5.
Now changing E postion will give me one extra R1E2E1R2R3 of the (form REERR)

Total duplicates or extra =6 for each unqiue arrangement of the word REERR
FOR arrangmnt of REERR there are 6 extra .... and of EERRR 6 extra and so on...


Total arrangements of REERR with repetions=npr=5factorial

5 factrial consists of each unique arrangements {x1,x2...Xn}+6 duplicates of every each unique arrangements 6{x1,x2...Xn}={x1,x2...Xn}+6{x1,x2...Xn}

=>{x1,x2...Xn}+6{x1,x2...Xn}=5 factorial
=>7{x1,x2...Xn}=5 factorial
=>x1...Xn=5 factorail/7
WHy wrong? Why the previous method does not work in case2

@Dr.Peterson @Subhotosh Khan
 
Your method is not wrong, but in case two you made a mistake.
In case one you prooved that given a set , there are five duplicates. In case 2, given a set there are not 6 but 11 duplicates so the number of possibilities is 5 factorail/12= 10

There is an easy way to solve anagram problems(finding the number of way to re-arrange letters):

Let be the word W=X1X2X3X4...Xn and A={Y1,Y2,...,Ym} the set of letters of W with no redundancy

The number of anagrams of W(number of way to re-arrange W's letters) is:
n!/P
Where P=[(number of repeat of Y1)!*(number of repeat of Y2)!*...(number of repeat of Ym)!= number of repeat of each distinct anagram
(where n! denotes n factorial)

Let's apply it:

In case one
W=KELLLY and A={K,E,L,Y}

so the number of anagrams of W=6!/(1!*1!*3!*1!)=120

In case two:

W=REERR and A={R,E}

so the number of anagrams of W = 5!/2!*3!=5!/12=10

The explanation of the method is quite intuitive. Using the notations above, given an anagram of the word W, to have a duplicate of W we just have to permute positions of same letters and do the same for the others, given a letter Yk where k is between 1 and m, the number of permutation of positions of Yk inside the word W is (the number of repetition of Yk)!. So the number of duplicates of an anagram of W is (the number of repeat of Y1)!*(the number of repeat of Y2)!...(the number of repeat of Ym)!
N.B: we multiply those numbers because of and, in combinatory and is translated as a product.
 
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