How many 5 letter palindromes can be made using an alphabet with 5 letters?

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SUMMARY

The discussion centers on calculating the number of 5-letter palindromes that can be formed using an alphabet of 5 letters. The formula derived indicates that for an alphabet of size 4, the number of 5-letter palindromes is calculated using the pattern xyzyx, resulting in 64 unique combinations. The Python 3.1 code provided demonstrates how to generate these palindromes programmatically. The conversation also clarifies that the number of palindromes increases with the length of the alphabet and the word length.

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Maria76
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Hi,

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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Hi Maria! :smile:

This should help you to find the formula …

If the word-length is 2n (even), then the number of palindromes is the same as the number of ordinary words of length n.

If the word-length is 2n - 1 (odd), then the number of palindromes is the same as the number of palindromes of length 2n, since you just double the central letter. :wink:
 
Maria76 said:
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)?

You need to add some limit to the word length. Answer to the question as stated is "infinitely many". Take any, add the same letter at the end and at the beginning, and you have a new palindrome. Repeat ad nauseam.
 
Maria76 said:
Hi,

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

Why do you also use the letters E and F? I thought your alphabet consists only of the letters A,B,C,D.

To find the number of combinations write down what a palindrome is in general:
xyyx
 
Edgardo said:
Why do you also use the letters E and F? I thought your alphabet consists only of the letters A,B,C,D.

To find the number of combinations write down what a palindrome is in general:
xyyx


Additionally, your examples are only 4 letters, not 5.

As shown above, you just need the pattern xy and pair it with it's reverse to get xyyx.

How many xy patterns are there? That's the Cartesian Product of 4 letters taken 2 at a time.

That would be (length of alphabet) * (length of alphabet).
(Or in general, (length of alphabet)**(length of pattern).)

The number of xy patterns is (4*4).

To get 5 letter palindromes, you need to use the pattern xyzyx, so you would need to multiply the number of xy patterns by the length of the alphabet.

Thus, the number of xyzyx patterns is (4*4)*4.

Which can be generated thusly:
Code: 
# Python 3.1
import itertools as it
count = 0
alpha = 'abcd'
for h in alpha:
  for i in it.product(alpha,alpha):
    j = ''.join(i)
    k = ''.join(reversed(j))
    print(j+h+k)
    count += 1
print(count)

aaaaa ababa acaca adada baaab bbabb bcacb bdadb caaac cbabc ccacc cdadc daaad dbabd dcacd ddadd aabaa abbba acbca adbda babab bbbbb bcbcb bdbdb cabac cbbbc ccbcc cdbdc dabad dbbbd dcbcd ddbdd aacaa abcba accca adcda bacab bbcbb bcccb bdcdb cacac cbcbc ccccc cdcdc dacad dbcbd dcccd ddcdd aadaa abdba acdca addda badab bbdbb bcdcb bdddb cadac cbdbc ccdcc cdddc dadad dbdbd dcdcd ddddd
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