Combination?

There are 4 consonants and 3 vowels.

Combine 2 of the consonants together in some order.

There are 4*3 = 12 ways to do this.

Now we have 3 'consonants' and 3 vowels

None of these 'consonants' can be placed next to each other.

Place the 3 vowels in some order: 3! = 6 ways to do this

Now we place the 3 'consonants' in one of the 4 positions (depicted with |) between the vowels or to the left or right of vowels: | v | v | v |

There are P(4,3) = 24 ways to do this

12 * 6 * 24 = 1728

It is difficult! Let C be a consonant and V be a vowel. There are 4 C's and 3 V's.

If the 2 consecutive consonants are in positions 1 and 2, then we have CCVCVCV, which occurs in 4*3*3*2*2*1*1 = 144 ways.

The 2 consecutive consonants may be in positions 12, or 23, or 34, or 45, or 56, or 67. So the 144 gets multiplied by 6 to equal 864.

I am off by a factor of 2 here, and I don't see why we should multiply by 2.

The 4 consonants can be arranged in 4! = 24 ways.

There are 3 spaces between them. Two must be filled with vowels. There are 3C2 = 3 ways to choose those.

The 3rd vowel must be placed at an end, or in the same place as one of the other two vowels. There are 4 choices for that.

And the 3 vowels can be arranged in 3! = 6 ways.

24 x 3 x 4 x 6 =1728