Using a 26-letter alphabet, how many 3-letter palindromes are there? How many 5-letter palindromes?
(A palindrome reads the same forwards and backwards, e.g. 'aba' or 'level'.)
How to Think About It: A palindrome is determined entirely by its first half -- the back half is forced to mirror the front. So just count the free positions and raise 26 to that power.
Quick Estimate: For 3 letters, the middle letter is free and the first determines the last, so two free choices: 6^2$. For 5 letters, the first three positions are free and the last two mirror them: 6^3$.
Formal Solution: A length-$n$ palindrome over an alphabet of size 6$ is fixed by its first $\lceil n/2 \rceil$ characters. - For $n = 3$: free positions are 1 and 2 (position 3 mirrors position 1), so 6^2 = 676$. - For $n = 5$: free positions are 1, 2, 3 (positions 4 and 5 mirror 2 and 1), so 6^3 = 17{,}576$.
Answer: $676$ three-letter palindromes and
This is a pure mental-math counting check, and the entire trick is realizing that the mirror constraint removes degrees of freedom. The count is
The same degrees-of-freedom counting underlies more serious work: counting the free parameters of a symmetric matrix, the dimension of a constrained space, or the number of independent entries after imposing structure. Whenever a symmetry or constraint ties some choices to others, the right count is over the free choices only.