How many six-letter words can be formed with the letters of the word ‘policy’ such that the vowels can only occur in even positions?

Question

We have $3$ places for evens and two vowels so selecting $2$ vowels for $3$ positions would be $3P2 =6$ but how can I do the same thing for consonants. Now I've left with $3$ places and $4$ consonants.

Answer 1

You have $3\cdot2 = 6$ possibilities for the vowels, and then with $4$ remaining places (the odd places plus the unfilled even place) you have $4 \cdot 3 \cdot 2 \cdot 1 = 24$ possibilities for the consonants. So in total there are $6 \cdot 24 = 144$ possible words.

Answer 2

First insert your vowels, as you did. Note that there is a 'blank' vowel among the places you're going to fill, as you only have two vowels for three spaces.

Then you're left with four spaces, as Henry already stated - the odd spaces and the even space you didn't use for the vowels.

So in conclusion, you have 6*24 permutations.