Pigeonholes problem help please

Question

Let $L$ be a list (not necessarily in alphabetical order) of the 26 letters in the English alphabet (which consists of 5 vowels, and 21 consonants).

a) Show that $L$ has a sublist consisting of four or more consecutive consonants.

b) Assuming that $L$ begins with a vowel, say A, show that $L$ has a sublist consisting of five or more consecutive consonants.

Answer 1

Assume there isn't a sublist that consists of consonants only and has a length of 4 or more. That means that we have a pattern like

XXX A XXX E XXX I XXX O XXX U XXX

with the 5 vowels and the rest are consonants. But with this pattern we've only used 18 out of 21 consonants. Adding one of the left over consonants somewhere in the list will always result in a sequence of 4 consonants.

If $L$ starts with a vowel, say A, we have a sublist, I call it $L'$, that consists of the 25 letters of the alphabet without an A. So we have 4 vowels and 21 consonants. Just like above they can be arranged like

XXXX E XXXX I XXXX O XXXX U XXXX

But again we have consonants left over, a single one in this case. And no matter where it is put into the list, it will make a sequence of 5 consecutive consonants.

Since the sublist $L'$ must have a sequence of 5 consecutive consonants, so must $L$.

Answer 2

Let $v_1$ be the first vowel in the list, $v_2$ the second one and so on.

Let $n_0$ be the number of consonants before $v_1$, $n_1$ the number of consonants between $v_1$ and $v_2$; $n_2$, $n_3$, $n_4$ are defined similarly; $n_5$ is the number of consonants after $v_5$: $$ L=\underbrace{c\dots c}_{n_0}\; v_1 \; \underbrace{c\dots c}_{n_1}\; v_2 \; \underbrace{c\dots c}_{n_2}\; v_3 \; \underbrace{c\dots c}_{n_3}\; v_4 \; \underbrace{c\dots c}_{n_4}\; v_5 \; \underbrace{c\dots c}_{n_5} $$ Then $n_0+n_1+n_2+n_3+n_4+n_5=21$.

The second case corresponds to forcing $n_0=0$.

Can you go on?