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First of all, $\text{TRIANGLE}$ has $8$ distinct letters, $3$ of which are vowels($\text{I, A, E}$) and rest are consonants($\text{T, R, N, G, L}$).

While attempting this, I came up with the idea of putting alternate vowels and consonants not to group same types together.

So, I decided to form two 'batteries'. [$\text{V}$ stands for Vowels and $\text{C}$ stands for consonants.]

$$\text{V} \text{ C}\text{ V} \text{ C}\text{ V} $$

And,

$$\text{C} \text{ V}\text{ C} \text{ V}\text{ C}$$

If we count all the permutations and then add them up (Mutually Exclusive Events), we can get total number of permutations.

Now, For the first case,

$3$ vowels can be arranged in the $3$ spaces required in $3! = 6$ ways

From $5$ consonants, $2$ spaces can be filled with consonants in $^5P_2 = 20$ ways

One battery, $(8 - 3- 2) = 3$ letters to arrange.

Total number of permutations : $6 * 20 * 4! = 2880$.

In Second case,

From $3$ vowels, $2$ spaces can be filled with vowels in $^3P_2 = 6$ ways

From $5$ consonants, $3$ spaces can be filled with consonants in $^5P_3 = 60$ ways.

One battery, $(8 - 2- 3) = 3$ letters to arrange.

Total number of permutations : $6 * 60 * 4! = 8640$

So, Total number of permutations for the word $\text{TRIANGLE} = 2880 + 8640 = 11520$

Again, My answer is incorrect, according to my textbook. They report $14400$ is the correct answer.

So, what did I miss here now? Please elaborate, and I'll be happy with any sort of help. [Seriously, this morning is getting even more hectic for me]

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Your cases are not exhaustive. You have not considered cases in which the consonants and vowels do not alternate. Consider, for instance, the arrangement ATRINGLE.

The word TRIANGLE has five distinct consonants. We can arrange them in $5!$ ways. This creates six spaces in which we can place the three vowels.
$$\square C \square C \square C \square C \square C \square$$ where $C$ denotes a consonant. To separate the vowels, we must choose three of these six spaces in which to place a vowel. Since the three vowels are distinct, they can be arranged in the chosen places in $3!$ ways. Hence, the number of arrangements of the letters of the word TRIANGLE in which no two of the vowels are adjacent is $$5!\binom{6}{3}3!$$

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    $\begingroup$ Hmm, now I get it completely. $\endgroup$ – Soumalya Pramanik May 28 '19 at 1:44
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See you have $6$ spaces for $3$ vowels (including external ones also). So total number of ways in which you can choose $3$ of them are ${6\choose 3}=20$. The three spaces will be filled by 3different types of vowels so we need to multiply it's by $3!$. Hence total ways $=20×3!=120$ also the $5$ consonants are different hence they too can be arranged in $5!$ ways. Hence total number of ways$=20×3!×5!=14400$

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  • $\begingroup$ So, I need to consider the external spaces not to miss some combinations. $\endgroup$ – Soumalya Pramanik May 27 '19 at 6:34
  • $\begingroup$ Can you add a picture so I can visualise? $\endgroup$ – Soumalya Pramanik May 27 '19 at 6:36
  • $\begingroup$ The picture size is >2mb. The server doesn't allow to upload more than 2. You can ask your doubt directly. $\endgroup$ – Archis Welankar May 27 '19 at 6:49
  • $\begingroup$ Awwwwww.... Okay no Problem! $\endgroup$ – Soumalya Pramanik May 27 '19 at 6:50
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    $\begingroup$ @ArchisWelankar You can illustrate your method by typing $$\square C \square C \square C \square C \square C \square$$, which produces $$\square C \square C \square C \square C \square C \square$$ or by typing $$\_ C \_ C \_ C \_ C \_ C \_$$, which produces $$\_ C \_ C \_ C \_ C \_ C \_$$ $\endgroup$ – N. F. Taussig May 27 '19 at 9:45

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