Answer is 576.
here are seven letters in the word ‘FAILURE’ out of which the consonants are- F, L and R. And the vowels are A,E,I and U. There are total 4 odd positions (1st, 3rd, 5th and 7th) and 3 even positions (2nd, 4th and 6th) to fill.
The constraint is that consonants may occupy only an odd position. So, first we’ll fill the odd positions with the consonants. There are total 4 odd positions and 3 consonants. Since there are only 3 consonants, then at a time, 3 consonants can occupy only 3 odd positions and one will be left out. So, we’ll need to select which three odd positions out of the four available odd positions we will fill first. The number of ways of selecting 3 odd positions out of four are:
This gives us 4 ways to select 3 odd positions out of the available 4 odd positions. Now that we have selected the odd positions to fill, we now need to arrange the consonants in these positions. The number of ways in which we can arrange 3 consonants is equal to 3.Now only the vowels remain to be arranged. There are 4 vowels and 4 vowels can be arranged in 4!.
Now we multiply these results to arrive at the answer: 4∗3!∗4!=576 ways.