In how many ways can u arrange the letters "No Ideas Help me" with given condition The N and O are NOT next to each other , the I is directly before the D and the letters EEE are NOT together( 2 E's can be together).
Here's what i tried to do but got stuck
N O together A= 12!2!/ 3! I Before D B= 12!/ 3! EEE together C= 11!
n= A + B + C - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C|
then i tried to change |B| since we need its complement to do Total - n
so |B| = 13!-12!\over 3!
Then when i try to do A ∩ B it gets complicated. I just need to know if theres an easier way of approaching this.
 A: Let's break this problem down.
First let's think about our $E$'s. When we put all of our $E$'s down we see that there are four slots created _$E$_$E$_$E$_ and we can fix our $E$'s here since the order doesn't matter (since each $E$ is the same). Now we have two slots that must have at least one element in them so that no $E$ is next to another $E$. We label our slots $(x_1)E(x_2)E(x_3)E(x_4)$ and treat this as an integer solution involving our remaining letters. That is $x_1+x_2+x_3+x_4=10$ with $x_2 \geq 1$ and $x_3\geq 1$. This gives $x_1+x_2\prime + x_3\prime + x_4 = 8$. By stars and bars there are thus $\binom{4-1+8}{8}$ ways to place our remaining elements. But before we place our remaining elements we must arrange them.
Let's look at how we can arrange our remaining letters without our $E$'s. We have $10!$ total arrangements of our remaining letters. For those where $N$ and $O$ are together we have $9!\cdot 2!$. We see that half of these permutations will have $D$ before $I$, and the other half will have $I$ before $D$ by symmetry. Thus we subtract out all arrangements with $N$ and $O$ together or with $D$ before $I$.
Thus our solution is
$$\left(10!-\left(9!\cdot 2! + 10! \cdot \frac{1}{2} - 9!\cdot 2! \cdot \frac{1}{2}\right)\right)\cdot \binom{11}{8}.$$
A: Since the $I$ is directly before the $D$, we can regard those two as a single element $\fbox{$ID$}$ throughout.
Then there are 12 elements, 3 identical, giving $\frac{12!}{3!}$ basic permutations. However a certain number are forbidden, which are those containing $\fbox{$NO$}$, $\fbox{$ON$}$ and $\fbox{$EEE$}$. These arise in $\frac{11!}{3!}$, $\frac{11!}{3!}$ and $10!$ ways, with $2\cdot9!$ arrangements having both problems. 
By inclusion-exclusion we get the final count of 
$$\frac{12!}{3!} - 2\frac{11!}{3!} - 10! + 2\cdot9!$$ 
