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bintang
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how many different strings over uppercase alphabet (A to Z) of length 9 are not palindromes?
 one year ago
 one year ago
bintang Group Title
how many different strings over uppercase alphabet (A to Z) of length 9 are not palindromes?
 one year ago
 one year ago

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mathmate Group TitleBest ResponseYou've already chosen the best response.1
Number of strings = 26^9 Number of palindromes = 26^5 Number of strings that are not palindromes = 26^926^5=5429491797600
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.0
i think to find the palindromic numbers, we must divided by some cases : case I : it's forms be AAAA(A)AAAA (next i always signed for the middle term) so, 4!/4! * 26P1 = 26P1 = 26!/25! = 26 case II : it's forms be AAAA(B)AAAA so, 4!/4! * 26P2 = 26P2 = 26!/24! = 26 * 25 = 650 case III : it's forms be AAAB(C)BAAA so, 4!/3! * 26P3 = 4 * 26!/23! = 4 * 26 * 25 * 24 = 62400 case IV : it's forms be AABC(D)CBAA so, 4!/2! * 26P4 = 12 * 26!/22! = 12*26*25*24*23 = 4305600 vase V : it's forms be ABCD(E)DCBA so, 4! * 26P5 = 24 * 26!/21! = 24*26*25*24*23*22 = 189446400 thus, the total of palindromic numbers = 26+650+62400+4305600+189446400 = 193815076 and we know that for all Number of strings = 26^9 therefore, the total different strings over uppercase alphabet (A to Z) of length 9 are not palindromes = 26^9  193815076 CMIIW (correct me if im wrong) :)
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.0
@shubhamsrg , @hba , @abb0t , @hartnn
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.0
please, verify...
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
seems to me that u covered all cases, but still take 2nd opinion ...
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.0
yeah, i hope anyone can give the other ways
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
please correct me if am wrong anywhere in the logic.. ____ _ ____ no. of ways to fill in first 4 spaces = 26^4 after filing of 1st 4 spaces, so as to form a palindrome, we have only 1 way to fill in last 4 for the 5th space, we can have 26 options, so total palindromes = 26^4 * 26 = 26^5 so somehow, i agree with @mathmate 's ans
 one year ago
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