Frank Day

2022-07-11

The letters of the word CONSTANTINOPLE are written on 14 cards, one of each card. The cards are shuffled and then arranged in a straight line. How many arrangements are there where no two vowels are next to each other?

Jenna Farmer

Beginner2022-07-12Added 17 answers

Step 1

First of all just consider the pattern of vowels and consonants.

We are given 5 vowels, which will split the sequence of 14 letters into 6 subsequences, the first before the first vowel, the second between the first and second vowels, etc.

The first and last of these 6 sequences of consonants may be empty, but the middle 4 must have at least one consonant in order to satisfy the condition that no two vowels are adjacent.

That leaves us with 5 consonants to divide among the 6 sequences. The possible clusterings are $\left\{5\right\}$, $\{4,1\}$, $\{3,2\}$, $\{3,1,1\}$, $\{2,2,1\}$, $\{2,1,1,1\}$, $\{1,1,1,1,1\}$.

The number of different ways to allocate the parts of the cluster among the 6 subsequences for each of these clusterings is as follows:

$\left\{5\right\}:6$

$\{4,1\}:6\times 5=30$

$\{3,2\}:6\times 5=30$

$\{3,1,1\}:\frac{6\times 5\times 4}{2}=60$

$\{2,2,1\}:\frac{6\times 5\times 4}{2}=60$

$\{2,1,1,1\}:\frac{6\times 5\times 4\times 3}{3!}=60$

$\{1,1,1,1,1\}:6$

Step 2

That is a total of 252 ways to divide 5 consonants among 6 subsequences.

Next look at the subsequences of vowels and consonants in the arrangements:

The 5 vowels can be ordered in $\frac{5!}{2!}=60$ ways since there are 2O's.

The 9 consonants can be ordered in $\frac{9!}{3!2!}=30240$ ways since there are 3 N's and 2T's

So the total possible number of arrangements satisfying the conditions is $252\cdot 60\cdot 30240=457228800$.

First of all just consider the pattern of vowels and consonants.

We are given 5 vowels, which will split the sequence of 14 letters into 6 subsequences, the first before the first vowel, the second between the first and second vowels, etc.

The first and last of these 6 sequences of consonants may be empty, but the middle 4 must have at least one consonant in order to satisfy the condition that no two vowels are adjacent.

That leaves us with 5 consonants to divide among the 6 sequences. The possible clusterings are $\left\{5\right\}$, $\{4,1\}$, $\{3,2\}$, $\{3,1,1\}$, $\{2,2,1\}$, $\{2,1,1,1\}$, $\{1,1,1,1,1\}$.

The number of different ways to allocate the parts of the cluster among the 6 subsequences for each of these clusterings is as follows:

$\left\{5\right\}:6$

$\{4,1\}:6\times 5=30$

$\{3,2\}:6\times 5=30$

$\{3,1,1\}:\frac{6\times 5\times 4}{2}=60$

$\{2,2,1\}:\frac{6\times 5\times 4}{2}=60$

$\{2,1,1,1\}:\frac{6\times 5\times 4\times 3}{3!}=60$

$\{1,1,1,1,1\}:6$

Step 2

That is a total of 252 ways to divide 5 consonants among 6 subsequences.

Next look at the subsequences of vowels and consonants in the arrangements:

The 5 vowels can be ordered in $\frac{5!}{2!}=60$ ways since there are 2O's.

The 9 consonants can be ordered in $\frac{9!}{3!2!}=30240$ ways since there are 3 N's and 2T's

So the total possible number of arrangements satisfying the conditions is $252\cdot 60\cdot 30240=457228800$.

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