anonymous
  • anonymous
how many different strings over uppercase alphabet (A to Z) of length 9 are not palindromes?
Probability
schrodinger
  • schrodinger
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mathmate
  • mathmate
Number of strings = 26^9 Number of palindromes = 26^5 Number of strings that are not palindromes = 26^9-26^5=5429491797600
RadEn
  • RadEn
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) :)
RadEn
  • RadEn

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RadEn
  • RadEn
please, verify...
hartnn
  • hartnn
seems to me that u covered all cases, but still take 2nd opinion ...
RadEn
  • RadEn
yeah, i hope anyone can give the other ways
shubhamsrg
  • shubhamsrg
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
anonymous
  • anonymous
thanks :)

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