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

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Discussion Overview

The discussion revolves around the number of 5-letter palindromes that can be formed using an alphabet of 5 letters. Participants explore the concept of palindromes, provide examples, and consider mathematical approaches to calculate the total combinations.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Maria initially counts 36 palindromic combinations using an alphabet of 4 letters, providing specific examples.
  • One participant suggests a formula for calculating palindromes based on word length, indicating that odd-length palindromes can be derived from even-length ones.
  • Another participant argues that without a limit on word length, there are infinitely many palindromes, as any palindrome can be extended indefinitely.
  • There is confusion regarding the use of letters E and F in examples, as some participants assert the alphabet consists only of A, B, C, and D.
  • A later reply clarifies the structure of palindromes and proposes a method to calculate the number of 5-letter palindromes using patterns and Cartesian products.

Areas of Agreement / Disagreement

Participants express differing views on the number of palindromic combinations possible, with some asserting a finite count and others suggesting infinite possibilities. There is also disagreement regarding the correct alphabet used in examples.

Contextual Notes

Some participants' calculations depend on the definitions of palindromes and the constraints of the alphabet, leading to unresolved questions about the total number of combinations.

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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