Number of words in which all the vowels are not together of the word GANESHPURI? 
Number of words in which all the vowels are not together of the word GANESHPURI?

The options available are 


*

*$21\cdot 7!$

*$42\cdot 8!$

*$84\cdot 7!$

*None


I have found words with vowels always together to be equal to $7!\cdot 4!$ and subtracted total number of words from it $(10!-7!×4!)$ but it seems wrong. 
Also;
2:) Number of words with any two of the letters E,H and  P are never together?
 A: To start with we will have
$$
V\ C\ V\ C\ V\ C\ V
$$
since we have $4$ vowels and we have to have a consonant inbetween each vowel. For this we have to pick $3$ consonants out of $6$ which we can do in
$$
\binom{6}{3}
$$
different ways and we can place them in these spots in $3!$ different ways so
$$
\binom{6}{3}\cdot 3!.
$$
The four vowels we can place in $4!$ different permutations into the said spots sowe have
$$
\binom{6}{3}\cdot 3!\cdot 4!
$$
different combinations so far and we have to put the rest of the $3$ consonants in any place we like. There are $8$ spots for the firts one and after we placed this one we will have $9$ spots for the second one and finally $10$ spots for the last onegiving
$$
\binom{6}{3}\cdot 3!\cdot 4!\cdot 8\cdot 9\cdot 10=4\cdot 8\cdot 9\cdot  10\cdot 6!=2880\cdot 6!
$$
And none of the alternatives given is this number.

For the second question you can count the number of ways putting these letters together in any combination and subtract it from all possibilities.
So I would say that first we clump these up and handle as one entity, there are $10$ different letters where we clump the given $3$ together so we have $8$ objects to order (seven letters and a clump) which we can do in 
$$
8!
$$
different ways. Now inside the clump we can arrange these letters in $3!$ different ways so we have
$$
8!\cdot3!
$$
so far.
Can you continue?

EDIT
After reading the second question I realised that I answered the first question in a wrong way, I counted the number of ways where NONE of the vowels are next to eachohther.
Your approach was correct 
$$
10!-7!\cdot 4!=7!(8\cdot 9\cdot 10-4!)=7!\cdot 696
$$

EDIT after further question in a comment
So when you are done assigning letters to the pattern
$$
V\ C\ V\ C\ V\ C\ V
$$
you made sure that there are no vowels together so where you put the remaining $3$ consonants does not matter. I put an $O$ to the possible places for the first of these three
$$
O\ V\ O\ C\ O\ V\ O\ C\ O\ V\ O\ C\ O\ V\ O
$$ 
these are $8$ different spots. When you chose to put it somewhere (for the examples sake I put it somewhere) 
$$
V\ C\ V\ C\ C\ V\ C\ V
$$
Now you have $9$ spots for the next one, and so on.
