Adeel,

First off this seems like a data sufficiency question which means that you dont have to determine a final answer but merely sufficiency. That being sad i think stmnt 1 is insufficient by itself given the existence of statement two or in other words you cannot be sure of any other constraints (except M is to remain put in the third position).

Likewise stmnt two is insufficient as well. Hence the answer choice drills down to either C or E.

I came up with 144 combinations and C as an answer choice by thinking about it like this:

C-O-M-P-U-T-E-R (8 slots minus 1 since M cant be moved so 7 slots)

now for the first slot i have 4 possibilities (CPTR) since vowels must be in the even slots. Furthermore the next slot i dealt with was slot no. 5 (U) here i have 3 possibilities (since 1 has been used up in the first). continuing this methodology the odd slots (excluding M) can be arranged in 4*3*2*1 or 24 ways. we exlude M since it cant be moved.

The even slots can be arranged in 3*2*1 (or 6 ways). The total combinations thus are 24*6 or 144.

Hope this helps. Thanks.

I though 1 & 2 were both required at the same time. Here is what I got:
4C3 {4 even position, 3 vowels} * 1 {M is 3rd} * 4! = 96

I bet this question is not from official guides/prep, because it's stated poorly.