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I unfortunately don't have enough reputation to comment, forgive the answer that is a link to another answer. Your question is explained well in this answer: http://crypto.stackexchange.com/a/12706/17884


It's not elegant, but the brute force method is to write a program that creates a table of 25x25 digraphs (assuming i=j), yielding 625 rows. I'd also add a column that lists the relative frequency of each digraph (given enough ciphertext you can use that to identify frequent substitutions, as you already have done). You start off with 625! possible ...

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