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Here is a weird question (I hope you don't mind).

Let's say we have an alphabet with only 4 letters (A, B, C, D)? How many combinations of give us palindromes (i.e. they read the same way backwards and forwards)? I count 36 possible combinations -

AAAA BAAB CAAC DAAD EAAE FAAF

ABBA BBBB CBBC DBBD EBBE FBBF

ACCA BCCB CCCC DCCD ECCE FCCF

ADDA BDDB CDDC DDDD EDDE FDDF

AEEA BEEB CEEC DEED EEEE FEEF

AFFA BFFB CFFC DFFD EFFE FFFF

This is the same number with an alphabet of only 3 letters -.

AFA, AEA, ADA, ACA, ABA, AAA

BFB, BEB, BDB, BCB, BBB, BAB

CFC, CEC, CDC, CCC, CBC, CAC

DFD, DED, DDD, DCD, DBD, DAD

EFE, EEE, EDE, ECE, EBE, EAE

FFF, FEF, FDF, FCF, FBF, FAF

Is that right?

Now, how about an alphabet with 5 letters? Is there any equation that can be used to determine the number of palindromes, or do I need to use a computer algorithm to work it out.

Thanks,

Maria

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# Palindrome with 5 letters

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