Question

how many different strings over uppercase alphabet (A to Z) of length 9 are not palindromes?

Answer 1

Number of strings = 26^9
Number of palindromes = 26^5
Number of strings that are not palindromes = 26^9-26^5=5429491797600

Answer 2

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) :)

Answer 3

@shubhamsrg , @hba , @abb0t , @hartnn

Answer 4

please, verify...

Answer 5

seems to me that u covered all cases, but still take 2nd opinion ...

Answer 6

yeah, i hope anyone can give the other ways

Answer 7

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

Answer 8

thanks :)