How many $4$-permutations are there of the set $\{A,B,C,D,E,F\}$ if whenever $A$ appears in the permutation, it is followed by $E$? Case 1: when $A$ does not appear in the $4$-permutation
$5 \cdot 4 \cdot 3 \cdot 2 = 120$
Case 2: when $A$ does appear in the $4$-permutation. Then $E$ should follow $A$. Let $X$ stand for $AE$, now we should consider $3$-permutations, since $X$ occupies two spaces. Since $X$ stands for two elements, we consider the number of elements to be $5$.
$5 \cdot 4 \cdot 3 = 60$
Thus, number of permutations of the set $\{A,B,C,D,E,F\}$ where if whenever $A$ appears in the permutation, it is followed by $E$ is $120 + 60 = 180$.
Is my reasoning correct?
EDIT
Alternative reasoning for Case 2
when A does appear in the 4-permutation. Then E should follow A
Let X stand for AE, now we should consider 3-permutations, since X occupies two spaces
Now X can be arranged in the following way

*

*X, *, *

**, X, *

**, *, X

Thus we have

*

*$1 \cdot 4 \cdot 3 = 12$

*$4 \cdot 1 \cdot 3 = 12$

*$4 \cdot 3 \cdot 1 = 12$
So for case 2 the total number of permutations = $12 + 12 + 12 = 36$
Thus  Number of permutations of the set {A,B,C,D,E,F} if whenever A appears in the permutation, it is followed by E is = $120 + 36 = 156$
So which of my reasoning is correct? And why is the other one incorrect?
Help greatly appritiated
Thanks
 A: You handled case 1 correctly.  Your second approach to case 2 is correct.  In your first approach to case 2, there are not five elements left to choose once you include AE, just two of the remaining four elements.
For case 2, if $E$ immediately follows $A$, there are three places to place the block $AE$ since it must begin in one of the first three positions, four ways to place one of the remaining four letters in the first open position, and three ways to place one of the remaining three letters in the last open position.  Hence, there are $3 \cdot 4 \cdot 3 = 36$ admissible arrangements in case 2.
Alternatively, there are $\binom{4}{2}$ ways to select the other two letters which appear with the block $AE$ and $3!$ ways to arrange the block $AE$ and the other two selected letters, so there are $\binom{4}{2}3! = 36$ admissible permutations in case 2.
Hence, there are a total of $120 + 36 = 156$ admissible $4$-permutations of the letters $\{A, B, C, D, E, F\}$ if whenever $A$ appears in the permutation, it is followed by $E$.
