In how many ways can the letters of the word MANAGEMENT be arranged such that no two vowels appear together?

MANAGEMENT has 2 M's, 2 A's, 2 N's, 1 G, 2 E's, 1 T

Vowels are A and E.

No two vowels are together. So, between every vowel there has to be atleast a consonant.

So, if we write the 6 consonants like:

$$\times\ M\times\ M\times\ N\times\ N\times\ G\times\ T\times\ $$

There are 7 gaps, out of which we need to place 4 vowels.

No. of ways of selecting 4 places out of 7 and then placing the 4 vowels are $$^7C_4\times\ \dfrac{4!}{2!\cdot2!}$$

Number of ways in which the 6 consonants can be arranged =$$\dfrac{6!}{2!\cdot2!}$$

So, total number of permutations =$$^7C_4\times\ \dfrac{4!}{2!\cdot2!}\times\ \dfrac{6!}{2!\cdot2!}$$

=$$\dfrac{7!}{4!\cdot3!}\times\ \dfrac{4!}{2!\cdot2!}\times\ \dfrac{6!}{2!\cdot2!}=\dfrac{7!}{3!}\times\ \dfrac{6!}{\left(2!\right)^4}=\dfrac{5040}{6}\times\ \dfrac{720}{16}=315\times\ 120=37800$$