Permutation with no two vowels next to each other

Question

Question : find number of arrangements of the word TRIANGLE in which no two vowels are next to each other.

My attempt : $5! ( ^6P_3) =14,400$

Is this correct?

Answer 1

Yes, your answer is correct.

There are $5$ consonants and $3$ vowels.

Vowels can be selected in $\dbinom{6}{3}$ ways.

The vowels be arranged in $3!$ ways

The consonants can be arranged in $5!$ ways.

In total we have $\dbinom{6}{3}\times3!\times5!=14400$ ways.

Answer 2

The consonants can be arranged in $5!$ ways. There are then six "slots" available for single vowels: one slot on each side of the five consonants:

_ C _ C _ C _ C _ C _

You can choose which of those six slots to fill in ${6 \choose 3}$ ways. And for each such choice there are $3!$ ways to put the three vowels.

Thus: $5! \times (6 \cdot 5 \cdot 4)/(3 \cdot 2 \cdot 1) \times 3! = 14400 $ ways.