Permutations of MULTIPLE

Question

The number of distinct rearrangements of the word "MULTIPLE" that can be made preserving the order in which the vowels (U,I,E) occur is what?

The answer given is $$\frac{8!}{(2!)6}$$ i dont get how they come to this answer

Answer 1

There are $8!$ possibilities if the two letters $L$ are taken to be distinct and no conditions are posed.

So there are $\frac{8!}{2!}$ possibilities if the two letters $L$ are not taken to be distinct. The double counting is repaired by dividing by $2$.

There are not $3!=6$ orderings but only one ordering for the vowels (since the order is fixed).

Also this must be repaired, this time by dividing by $3!=6$.

Final result: $$\frac{8!}{2!3!}=\frac{8!}{2\cdot6}$$

Answer 2

Out of $8$ positions, select $3$ positions for $U,I,E$ in $\binom83$ ways. $U,I,E$ will not permute but the remaining $5$ letters permute in $\displaystyle\frac{5!}{2!}$ ways, because $L$ repeats. The total permutations are $\displaystyle\binom83\cdot\frac{5!}{2!}=\frac{8!}{2!3!}$.