In "EAMCET", there are 3 vowels and 3 consonants. The consonants are all
distinguishable, but two of the vowels, the 2 E's, are indistinguishable.
First we calculate the total number of arrangements with no restrictions
other than that all the arrangements be distinguishable. That number would
be 6! if all letters were distinguishable, but we must divide by 2! because
of 2 indistinguishable E's. So if there were no restrictions other than
that they be indistinguishable, the answer would be 6!/2! = 720/2 = 360
However, there are 4 types of arrangements that are not allowed,
those in which no vowels are together. The number of these must be
subtracted from the 360. These types are:
1. VCVCVC
2. VCVCCV
3. VCCVCV
4. CVCVCV
where C stands for a consonant and V stands for a vowel.
For each of those 4 types, there are 3!=6 ways to arrange
the 3 consonants.
For each of those (4)(6)=24 ways to pick the types and
arrangements of the 3 consonants, there are 3!/2!=6/2=3
ways to choose the vowels. We divided by 2! because
there are 2 indistinguishable vowels, the E's.
So there are (4)(6)(3)=72 arrangements that are not allowed.
Answer: 360 - 72 = 288 ways
Edwin
