>>17721726>>17721726First consider all possible permutations of the non-vowels, I, N, D, P, N , D, N , C. There are 8 symbols, three N, two D, of permutations of these symbols is 8!/(3!*2!) = 3360.
Now the vowels must be together, so for each of these 3360 distinct permutations of these consonants, there are 9 places we could put the block of E's
So the answer is 3360*9 = 30240
Here's also a brute force programming solution (in Mathematica) based on generating all possible permutations and counting those that match the criterion
