Possible arrangments Letters? How many arrangements are possible of the letters in EZPZ I CAN DO IT, which has five vowels (A, E, I, I, O) and seven consonants (C, D, N, P, T, Z, Z).
a)  if there are no restrictions,  
b)  if the letters must be in alphabetical order?   
c)  if the vowels must be together?  
d)  6 letter words if the vowels must be together.
 A: (a) If there are no restrictions, we permute. However, since $I$ and $Z$ are repeated twice, we have to divide out according to the multinomial distribution:
$$\dfrac{12!}{2! * 2!}$$
(b) If you count the two $I$ and two $Z$ characters as the same, there is only one string with the letters in alphabetical order. If you count the two $I$ and two $Z$ characters as distinct, we can swap them, so we get $4$ ways. I would personally treat the $I$ and $Z$ characters as the same.
(c) Start by grouping the vowels. There are $\dfrac{5!}{2!}$ ways to permute the vowels. We then permute the consonants. There are $\dfrac{7!}{2!}$ ways to do this. You then place the vowel cluster in the consonants string. There are $8$ ways to do this. So by rule of product, we get:
$$\dfrac{5! * 7!}{2! * 2!} * 8 $$
(d) We again permute the vowels, which gives us $\dfrac{5!}{2!}$. We then choose a consonant. The $z$ is counted twice, so we have $6$ ways to choose a consonant. It must go at the beginning or end. There are two ways to do this. So by rule of product we get:
$$ \dfrac{5!}{2!} * 6 * 2 = 6! $$
