What's wrong with my permutation logic? The given question:

In how many ways the letters of the word RAINBOW be arranged, such that A is always before I and I is always before O.

I gave it a try and thought below:

Letters A, I and O should appear in that order. Then there are four places in which all the remaining four letters can appear. It means there are a total of 16 places: $$*_1*_2*_3*_4\quad A\quad*_5*_6*_7*_8\quad I\quad*_9*_{10}*_{11}*_{12}\quad O \quad *_{13}*_{14}*_{15}*_{16}$$ 
  Out of these 16 places remaining, the letter can appear in any of the 4 places, giving total $^{16}P_4 = 43680$ possible arrangements. 

But the given answer is just 840 with following explanation:

All the 7 letters of the word RAINBOW can be arranged in 7! ways and 3 particular letters (A,I,O) can be arranged in 3! ways. But the given condition is satisfied by only 1 out of 6 ways. Hence required number of arrangements 
  $$=\frac{7!}{3!} = 840$$

So what's wrong with my logic? I must be counting certain arrangements multiple times or my logic must be at fault in large. But what's that exactly?
 A: For the arrangement RAINBOW you have counted four places the R can be, four places the W can be, and six ways of putting NB in that order between the I and O, so you have counted it $96$ times. Other arrangements have been counted multiple times as well.
A: Ross Millikan has already explained the error in your reasoning. Here’s a correct version of the kind of calculation that you were trying to make. There are $\binom73$ ways to choose $3$ positions for the letters A, I, and O. The remaining $4$ letters can fill the remaining $4$ slots in any order, so they can be inserted in $4!$ ways. That gives you a total of
$$4!\binom73=\frac{7!}{3!}=840$$
arrangements.
A: Your mistake is here:

Out of these 16 places remaining letter can appear in any 4 places,
  giving total $^{16}P_4=43680$ possible arrangements.

The 43680 placements of letters into four of the 16 positions don’t yield different arrangements of the letters. For example, one arrangement is RAINBOW. Is the letter R in position 1, 2, 3, or 4? You counted each of those placements separately.
A: You would be entirely correct if there really were 16 places ( 19 in all ) and the places not filled with letters were filled with spaces so that
R***A****I*N*BO**W*

was truly distinct from 
*R**A****I*N*BO***W    ( I has used * because spaces are hard to count )
what $ ^{16} P _4$ counts is the number of four-tuples with non-repeating elements from 1 to 16. 
You might have the first element represent the position ( in your scheme of 16 positions ) of the letter "R" and the other three the positions of "N", "B", and "W" respectively.
so that  R***A****I*N*BO**W* would be represented as (1, 10, 12, 15)
and *R**A****I*N*BO***W would be represented as (2, 10, 12, 16) . 
A: I think the closest method to your original attempt is the following: after laying out the fixed letters A, I and O, consider one additional letter at a time.
(1) We have: " A I O ". R can be placed in any of four places.
(2) We now have four placed letters, for example " R A I O ". In any case we have five locations in which the next letter N can be placed.
(3) Similarly when we place the B, there are 6 choices of where to put it among the existing 5 letters.
(4) And finally the last letter W can be placed among the 6 existing letters in any of 7 places.
Each combination of choices will lead to a different outcome, as any two outcomes must differ by the order of as at least one pair of letters.
So the total number of orders is: 4 * 5 * 6 * 7 = 840
A: I got the answer above, but when I tried to solve the same problem I applied slightly different thinking to same solution.  Thus sharing it here.
The book says:

All the 7 letters of the word RAINBOW can be arranged in 7! ways and 3 particular letters (A,I,O) can be arranged in 3! ways. But the given condition is satisfied by only 1 out of 6 ways. Hence required number of arrangements 
  $$=\frac{7!}{3!} = 840$$

We can also think it as follows:

There are 7 letters in $RAINBOW$. Out which 3 are A, I and O, in that order (i.e. they cannot be in any other order I, A and O). So they can be at any of $\binom{7}{3}$ positions. Now remaining 4 letters are $R,N,B$ and $W$, in any order, so they take $4P_4$ positions. The total is
  $$\binom{7}{3}\times4P_4=\frac{7!}{4!\times3!}\times4!=\frac{7!}{3!}$$

Just a little bit of different thinking.
