No of vowels =4
no of ways of choosing vowels=4c2 ways
now you can interchange the two vowels therefore
total no of ways=4c2*2!=4P2=12

no of ways to arrange remaining words=5!

ans=1440=5!*12

that's correct rah_pandey.
I also solved it after I posted it:) But thought would be interesting to others maybe...


Though in my approach instead of this:
no of ways of choosing vowels=4c2 ways
now you can interchange the two vowels therefore
total no of ways=4c2*2!=4P2=12

I simply used the formula n!/[(n-k)!] which counts number of ways things can be arranged when order matters. then I multiplied it by 5! just like you, thus 1440

ket wrote:How many ways are there to rearrange the letters in the word 'elation', if the first and the last letter mus be each a vowel?

For the first letter you have 4 choices and for the last you have 3. So one example is

elatino

But if you reverse the position of the first and last you have a different arrangement:

olatine

After your first and last are decided, you will have 2 letters gone from 7 so only 5 letters to choose from for your middle letters. This is just 5! Putting all together we have:

4 x 5! x3 =1440