Getting two different answers for a permutation problem Q. Determine the number of ways the letters of the word $'TRIANGLE'$ can be arranged without placing the vowels side by side.
Approach $1$:
$5!\times\binom 63 \times3!=14,400$
Approach $2$:
Number  of  ways  the  letters  of  the  word  can  be  arranged  without  placing   the  vowels  side  by  side =  Number  of  ways  the  letters  of  the  word  can  be  arranged -  Number  of  ways  the  letters of the  word  can  be  arranged  when  placing  the  vowels  side  by  side
$= 8! - 6!3! = 36,000$
Could anyone please tell me why I'm getting different answers?
Thanks in advance!
 A: Your first approach is correct.  In approach 2, you subtracted those arrangements in which all three vowels are consecutive, but overlooked the possibility that exactly two of the vowels are adjacent.
Approach 1:  Arrange the five distinct consonants T, R, N, G, L in $5!$ ways.  This creates six spaces, four between consonants and two at the ends of the row, in which to insert the vowels.
$$\square C_1 \square C_2 \square C_3 \square C_4 \square C_5 \square$$
Choose three of the six spaces in which to place the vowels in $\binom{6}{3}$ ways, then arrange the vowels in the selected spaces in $3!$ ways.  This gives
$$5!\binom{6}{3}3! = 14,400$$
arrangements of the letters of the word TRIANGLE in which no two of the vowels are adjacent.
Approach 2:  There are $8!$ ways to arrange the letters of the word TRIANGLE. You subtracted the $6!3!$ cases in which all three vowels are consecutive.  However, we must also subtract the cases in which exactly two vowels are adjacent.
For the case in which exactly two of the three vowels are adjacent, arrange the five distinct consonants T, R, N, G, L in $5!$ ways.  This creates six spaces, four between consonants and two at the ends of the row, in which to insert the vowels.
$$\square C_1 \square C_2 \square C_3 \square C_4 \square C_5 \square$$
Choose one of these six spaces in which to place two vowels and one of the remaining five spaces in which to place the remaining vowel.  Finally, arrange the three vowels in the selected spaces from left to right.  There are
$$5! \cdot 6 \cdot 5 \cdot 3!$$
arrangements in which exactly two of the vowels are adjacent.
Hence, the number of admissible arrangements is
$$8! - 5! \cdot 6 \cdot 5 \cdot 3! - 6!3! = 14,400$$
Note:  Another of way of subtracting the cases in which at least two vowels are adjacent is to use the Inclusion-Exclusion Principle.  There are $8!$ ways to arrange the letters of the word TRIANGLE.
If two of the vowels are adjacent, we have seven objects to arrange, the block of two vowels and the other six letters.  There are $\binom{3}{2}$ ways to select the vowels in the block, $7!$ ways to arrange the seven objects, and $2!$ ways to arrange the vowels within the block.  However, if we subtract the
$$\binom{3}{2}7!2!$$
arrangements with a pair of adjacent vowels, we will subtract too much since we will have subtracted each case with two pairs of adjacent vowels twice, once for each way we could have designated one of those pairs as the pair of adjacent vowels.  Therefore, we have to add such arrangements to the total.
Since there are only three vowels, the only way to obtain two pairs of adjacent vowels is to have three consecutive vowels.  If all three vowels are consecutive, we have six objects to arrange, the five consonants and the block of three vowels.  The six objects can be arranged in $6!$ ways.  The three vowels can be arranged within the block in $3!$ ways.  Thus, there are
$$6!3!$$
arrangements in which all three vowels are consecutive.
By the Inclusion-Exclusion Principle, the number of admissible arrangements is
$$8! - \binom{3}{2}7!2! + 6!3! = 14,400$$
