permutations Problem

Additional Maths problem from text book.
Q15 Consider the letters of the word MONDAY. How many permutations of 4 different letters can be chosen if:
at least 1 vowel must be used?

Perms using vowel A and excluding O, I use 4P3=24 with 4 possible position for O makes 96
Same for vowel O excluding A, I get 4P3 = 24
with 4 possible positions for A makes 96

for 2 vowels permutation A O X X I use 4P2 to get 12 for the remaining non vowel letters and 6 possible arrangements for different position combinations of the 2 vowels. that makes another 96.
So 96+96+96 = 288. The book answer is 336.
Where am I going wrong?

Reply 1

mqb2766

Original post by xlaser31

Your 2 vowel case is undercounting the ways you can include the two vowels. You have 12 ways of arranging 2 consonants - then how may ways of arranging 2 consonants + 1 vowel? Then how many of 2 consonants + 2 vowels.

Note it would be a bit easier to consider the full set of permutations then subtract the number where there are no vowels. If the question asks for "at least 1 ..." then its often the simplest way to do it.