How many permutations can be made with the letters of the word ${}^{\prime }MEADOW{S}^{\prime }$ such that the vowels occupy even places?

$MEADOWS$

Consonant$=4$

Vowels$=3$

Places for vowels $=3!$ possibilities

Remaining consonants can be arranged in $4!$ Possibilities 

Total arrangements

$$\begin{gathered} 3!\times 4! \\ =6\times 24 \\ =144 \end{gathered}$$

Hence the correct answer is $144$