thanks for the explanations, gmat740 and venmuri. if it is assumed that any letter can be used only once in a word, then the explanation is quite clear.
but when i attempted the problem, i was also considering the possibility of any given alphabet appearing more than once in a word. in that case, we would have used a different formula and the answer would have been different than the given options.
if any vowel can appear more than once in a word, then using 4 vowels we can form 2^4=16 two-vowel sets. (aa, ee, ii, oo, ae, ai, ao, ea, ei, eo, ia, ie, io, oa, oe, oi)
similarly, if any consonant can appear more than once in a word, then using 8 consonants we can form 5^8=390625 five-consonant sets.
again, the 7 alphabets in a word can be arranged in 7! ways= 5040 ways.
so, the total number of possible arrangements= 16*390625*5040
let me admit that i am weak in permutation-problems. but i was thinking something along this line. if for a moment we forget the OA, can you please verify whether my approach is right? i have a feeling that, even if we consider that the vowels and consonants can repeat in a word, then also my answer is wrong :roll:
in short, can you please find out how many 7 letter combinations are possible from 4 vowels and 8 consonants, where any combination has 2 vowels and 5 consonants and any vowel or consonant can be repeated? (eg. aabbbbb is also permitted)
