How many different words can be formed from the letters of the word GANESHPURI when the vowels always occupy even places that is 2nd 4th etc?

Check out the new look and enjoy easier access to your favorite features

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.

12 A2 B2 C3456 A6 B6 C6 D7 A7 B7 C7 D7 E891011 A11 B11 C12