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Math Challenge II-A Combinatorics
Lecture 1, Problem 1.28
I got $26 \cdot 25^6-16 \cdot 25^5$, but the answer says that it's $26 \cdot 25^6-21 \cdot 20 \cdot 25^5$. The list of consonants is: b, c, d, f, g, h, j, k, l, m, n, p, q, r, s, t, v, w, x, y, z. The pairs of consecutive consonants are: (b, c), (c, d), (f, g), (g, h), (j, k), (k, l), (l, m), (m, n), (p, q), (q, r), (r, s), (s, t), (v, w), (w, x), (x, y), (y, z). There are 16 pairs, so the second part is $16 \cdot 25^5$. The first part is $26 \cdot 25^6$.
We have to be a little careful here with what the phrase "consecutive consonants" means here.
The idea (similar to no consecutive repetitions meaning the same letter cannot appear twice in a row) is that we do not want words that start with consonants as the first two letters. Therefore it is not just things like "pq" or "cd" that are not allowed, but also things like "pc" or "zb". This is where the $21\cdot 20$ ($21$ choices for first consonant, $20$ for second) comes from.
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