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Math Challenge II-A Combinatorics

 
 
LiangNeo的头像
Math Challenge II-A Combinatorics
LiangNeo - 2019年09月1日 Sunday 15:14
 

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 $26 \cdot 25^6$.

 
ProfessorAreteem的头像
Re: Math Challenge II-A Combinatorics
ProfessorAreteem - 2019年09月4日 Wednesday 13:06
 

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.