How many Arrangement of "AMAZED" letter E Positioned between two A's (Not necessarily Flanked) I considered 'AEA' as one letter so there are 4 letters which can be arranged in 4!=24 ways. But my sheet is telling its 120  How? Please HELP! & What is flanked meaning here?
 A: Hint what about "AMEAZD"? (the E is between the A's)
A: Here is a variant of Taussig's argument; in my view it is interesting to present it separately because it is in some sense more general.
You can divide all the anagrams of AMAZED into many boxes in this way: for each one in which EAA appear in this order, you put it in the same box with the ones in which these three letters are reordered as AEA and AAE, respectively, and all the rest of letters is in the same order/position.
This proves that the anagrams in which AEA are in this order are 1/3 of the total. The same reasoning applies to each word with two A, one E and an arbitrary number of other letters (possibly repeated).
A: \begin{array}{c|cccccc|c}
Case:1\to &A&\underline{4\;ways}&\underline{3\;ways}&\underline{2\;way}&\underline{1\;way}& A&=24 ways\\~\\
Case:2\to &A&\underline{3\;ways}&\underline{2\;ways}& \underline{1\;way}& A&\color{red}{\underline{3way}}&=18 ways\\
~\\
Case:3\to &A&\underline{2\;ways}&\underline{1\;way}&A& \color{red}{\underline{2\;way}}&\color{red}{\underline{3way}}&= 12 ways\\
~\\
Case:4\to &A&\underline{1\;way}&A&\color{red}{\underline{2\;ways}}& \color{red}{\underline{1\;way}}& \color{red}{\underline{3way}}&=6ways
\end{array}
I have listed all the possibilites of positions of $A$, $E$, $A$ here.
The red colour shows that $E$ cannot be there and $A$ represents place of  'A' and the '$\underline{\hspace{1cm}}$' giving the number of ways of arrangements in each case. Now, summing all the possiblities we have
$$60 \;ways$$
Repeating the same thing by interchanging the role of $A$, we will have one more 60 ways.
$$\text{so, the total ways}=60+60=120\text{ways}$$
A: Total permutations for the word AMAZED = 6!/2! = 360.
Out of 360 total permutations, only 1/3rd will satisfy our condition where E is between 2 As.
That is because when you permute AEA (3 ways), only 1 permutation exists where E is between 2 As.
Thus, total arrangements = 360/3 = 120
A: The word AMAZED has six letters.  Thus, there are six ways to place the M, five ways to place the Z, and four ways to place the D.  Once those three letters have been placed, there is one way to fill the remaining positions with the letters A, E, and A, namely place the E in the middle open position and place the A's in the remaining positions.  Therefore, there are $6 \cdot 5 \cdot 4 \cdot 1 = 120$ possible arrangements of the letters of the word AMAZED in which the E is positioned between the two A's.
Alternate Method:  There are $\binom{6}{3}$ ways to choose the positions where the letters A, E, and A will go, one way to place them in that order in the chosen positions, and $3!$ ways to place M, Z, and D in the three remaining positions.  Thus, there are 
$$\binom{6}{3} \cdot 3! = \frac{6!}{3!3!} \cdot 3! = \frac{6!}{3!} = 6 \cdot 5 \cdot 4 = 120$$ 
ways to arrange the letters of the word AMAZED so that the letter E is positioned between the two A's. 
A nice alternative to the methods presented above can be found in the solutions provided by Federico Poloni and Angad.
