Didn't get the answer.
You can visualize each vowel as having two arms to which consonants can be attached $-V-\;\;-V-\;\;-V-\;\;-V-$ There are 8 such arms,and for placing 7 consonants, 1 arm to be left, but if you leave out an arm between two vowels, one of the arms "disappears", so in effect there are only 5 ways of leaving out an arm, $$\text{ thus answer} = 5 \cdot \frac{7!}{2!2!} \cdot \frac{4!}{2!}$$ ps If $c$ is the # of consonants, and $v$ the # of vowels, the coefficient of 5 can be obtained as $$\sum_{k=c-v}^{v-1}\binom{v-1}k\binom{v+1-k}{c-2k}$$ |