Alphabetic Character String Permutations...With Restrictions. How many permutations exist in the string $ABCDEFG$, starting from the smallest possible combination if the only direction allowed is forward? For example, B is the smallest possible combination in the string $BDEF$. The only direction being forward, $BD, BE, BF$ are larger permutations, etc.
NOTE: You can't input a character more than once in the same permutation (e.g. $ABCDF$ is allowed but $ABCBDF$ is not. $CDEF$ is allowed on it's own, but $CDEFC$ is not).
Plus, how many key permutations will exist in the same string $A,B,C,D,E,F,G$ if you can move forward and backwards but aren't allowed to repeat a character in the same permutation (e.g. $ABCDF$ is allowed but $ABCBDF$ is not. $CDEF$ is allowed on it's own, but $CDEFC$ is not)?
 A: There are 128 choices. They follow and from this you can see how to count them.
A
B
C
D
E
F
G
A   B
A   C
A   D
A   E
A   F
A   G
B   C
B   D
B   E
B   F
B   G
C   D
C   E
C   F
C   G
D   E
D   F
D   G
E   F
E   G
F   G
A   B   C
A   B   D
A   B   E
A   B   F
A   B   G
A   C   D
A   C   E
A   C   F
A   C   G
A   D   E
A   D   F
A   D   G
A   E   F
A   E   G
A   F   G
B   C   D
B   C   E
B   C   F
B   C   G
B   D   E
B   D   F
B   D   G
B   E   F
B   E   G
B   F   G
C   D   E
C   D   F
C   D   G
C   E   F
C   E   G
C   F   G
D   E   F
D   E   G
D   F   G
E   F   G
A   B   C   D
A   B   C   E
A   B   C   F
A   B   C   G
A   B   D   E
A   B   D   F
A   B   D   G
A   B   E   F
A   B   E   G
A   B   F   G
A   C   D   E
A   C   D   F
A   C   D   G
A   C   E   F
A   C   E   G
A   C   F   G
A   D   E   F
A   D   E   G
A   D   F   G
A   E   F   G
B   C   D   E
B   C   D   F
B   C   D   G
B   C   E   F
B   C   E   G
B   C   F   G
B   D   E   F
B   D   E   G
B   D   F   G
B   E   F   G
C   D   E   F
C   D   E   G
C   D   F   G
C   E   F   G
D   E   F   G
A   B   C   D   E
A   B   C   D   F
A   B   C   D   G
A   B   C   E   F
A   B   C   E   G
A   B   C   F   G
A   B   D   E   F
A   B   D   E   G
A   B   D   F   G
A   B   E   F   G
A   C   D   E   F
A   C   D   E   G
A   C   D   F   G
A   C   E   F   G
A   D   E   F   G
B   C   D   E   F
B   C   D   E   G
B   C   D   F   G
B   C   E   F   G
B   D   E   F   G
C   D   E   F   G
A   B   C   D   E   F
A   B   C   D   E   G
A   B   C   D   F   G
A   B   C   E   F   G
A   B   D   E   F   G
A   C   D   E   F   G
B   C   D   E   F   G
A   B   C   D   E   F   G
A: Ask yourself: how many exist for a 1-tuple, how many exist for a 2-tuple, etc. and then it should be easier for you to count.
A: Regarding the first question: for each position you have two options, on and off. So that gives a total of $2^7$ - 1. (The minus one excludes the empty case) 
The second question yields 7!. If it's not required to include all the letters then the answer is 7! + 7!+7!/2!+7!/3!+...+7!/6!
