Counting number of arrangements How many 8 length arrangements of word ABCDEFGH are possible with following constraints?


*

*ABCD and EFGH strings should come in the same order.You cannot permute ABCD and EFGH among themselves

*CD and EF should not interleave. It Means whenever CDEF comes in a particular arrangement then either it should be in order of CDEF or EFCD only.


My work:-
total ways these can be arranged (8!)/(4!*4!). But i am not able to adjust second constraint.
 A: This is my attempt. 
Let me first re-state your constraints, to ensure I understood them properly.
Let me use the notation $ N < M $ to state the letter $N$ appears before the letter $M$.
The first constraint then reads $ A < B < C < D$ and $ E < F < G < H$
The second constraint reads $ D < E $ OR $ F < C$, as a result of the fact that $C<D$ and $E < F$ following the first constraint. 
One starts observing that there is only one string such that $D < E$ (first case in the second constraint): ABCDEFGH. That possibility then is counted, we shall add it later.
We are left with the first constraint plus the constraint that $ F < C$.
The number of possibilities complying with such constraints is counted by considering the strings such that:
1)   $ A < B < C < D$ and $ E < F < G < H$ (first constraint) plus $H<C$ (all $ E, F, G, H$ before $C$)
2)  $ A < B < C < D$ and $ E < F < G < H$ (first constraint) plus  $G <C$ AND $H>D$ ($E, F, G$ before $C$, and $H$ after $C$)
3)  $ A < B < C < D$ and $ E < F < G < H$ (first constraint) plus  $F<C$  AND $G > D$ (only $E, F$ before $C$, and $G,H$ afterwards)
The number of possibilities for case is  found by a stars and bars approach, and equals
$ \binom{6}{4} $, $2 \binom{5}{3}  $ and $3 \binom{4}{2} $ respectively (the  factor multiplying the binomial coefficient coming from the number of ways in which the letters placed after $D$ can be placed in each case mentioned).
Summing it all, without forgetting the ABCDEFGH string mentioned at the beginning, one gets 54 possibilities in total.
A: Method 1:  We subtract those cases that fail to satisfy the second constraint from those that satisfy the first constraint.
If the second constraint were not a consideration, there would be $\binom{8}{4}$ ways of choosing the positions for A, B, C, D.  They must be placed in the chosen positions in that order.  The letters E, F, G, H must be placed in the remaining positions in that order.
From these, we must exclude those strings in which the letters C, D, E, F appear in one of the following orders:  CEFD, CEDF, ECDF, ECDF.  
If the letters CEFD appear in that order, the first two positions must be filled with A and B in that order.  Since G must follow F, D must appear in one of the last three positions.  Prohibited strings:  ABCEFDGH, ABCEFGDH, ABCEGHD.
If the letters CEDF appear in that order, the first two positions must be filled with A and B in that order and the last two positions must be filled with G and H in that order.  Prohibited string:  ABCEDFGH.
If the letters ECDF appear in that order, E must be in one of the first three positions since A and B must precede C and the last two positions must be filled with G and H in that order.  Prohibited strings:  ABECDFGH, AEBCDFGH, EABCDFGH.
If the letters ECFD appear in that order, E must in one of the first three positions since A and B must precede C and D must appear in one of the last three positions since G and H must follow F.  There are $3 \cdot 3 = 9$ such strings.  Prohibited strings:  ABECFDGH, ABECFGDH, ABECFGHD, AEBCFDGH, AEBCFGDH, AEBCFGHD, EABCFDGH, EABCFGDH, EABCFGHD.
Hence, the number of permissible strings is 
$$\binom{8}{4} - 3 - 1 - 3 - 9 = 54$$
Method 2:  We do a direct count, pivoting on the position of the C.
Since D must follow C, C cannot appear after the seventh position.  Since A and B must appear before C, C cannot appear before the third position.
Notice also that choosing the positions of A, B, C, and D completely determines the arrangement since the remaining positions must be filled by E, F, G, H in that order.
If C is in the seventh position, then D is in the last position and A and B must be in the first six positions.  Choosing the positions of A and B completely determines the string since they must appear in that order in the selected positions and E, F, G, H must appear in that order in the remaining positions.  The positions of A and B can be selected in $\binom{6}{2}$ ways.
If C is in the sixth position, it must be followed by D and H and preceded by A, B, E, F, and G.  Since D must appear in one of the last two positions and A and B must appear in two of the first five positions, the number of such strings is $\binom{2}{1}\binom{5}{2}$.
If C is in the fifth position, it must be followed by D, G, and H and preceded by A, B, E, and F.  Since D must appear in one of the last three positions and A and B must appear in two of the first four positions, the number of such strings is $\binom{3}{1}\binom{4}{2}$. 
It is not possible for C to appear in the fourth position without violating the second constraint.
If C is in the third position, then it must preceded by AB and followed by DEFGH, so ABCDEFGH is the only such string.  
Hence, the number of permissible strings is 
$$\binom{6}{2} + \binom{2}{1}\binom{5}{2} + \binom{3}{1}\binom{4}{2} + 1 = 15 + 2 \cdot 10 + 3 \cdot 6 + 1 = 15 + 20 + 18 + 1 = 54$$
as we found above.
A: If $CD$ comes before $EF$, then since $A$ and $B$ come before $C$, and $G$ and $H$ come after $F$, the order must be $ABCDEFGH$.
If $CD$ comes after $EF$, then since $G$ and $H$ must also come after $EF$, and $A$ and $B$ before $CD$, the first four letters must be $A,B,E,F$ with $A$ before $B$ and $E$ before $F$, which can be done in $\binom{4}{2} = 6$ ways. Similarly, the last four letters are $C,D,G,H$ with $C$ before $D$ and $G$ before $H$, which can be done in $6$ ways.
So there are a total of $1 + 6 \cdot 6 = 37$ valid strings.
A: Apart from the obvious $ABCDEFGH$, here is a way to count the ones with $EFCD$,
though there could be a slicker way (which I'm not able to see immediately)
$GH$ can be inserted in $6$ ways in $EFCD$, yielding $6$ substrings, viz.
$EFGHCD, EFGCHD, EFGCDH, EFCGHD, EFCGDH, \;and\; EFCDGH,$
and then $B$ and $A$ can be inserted as explained under for $EFGHCD$:  


*

*If $B$ is placed immediately to the left of $C$, starting with $EFGHBCD$, there are $5$ ways to insert $A; \;$ if $B$ is shifted one place to the left, there are $4$ ways for $A$, and so on down to just $1$ way for $A$ as $B$ is shifted left one by one, thus $5+4+3+2+1 = 15$ ways to insert $A$ and $B$

*For the next two substrings, similarly, there will be $2\times(4+3+2+1)=20$ ways,

*And for the next three substrings, there will be $3\times(3+2+1) = 18$ ways
Finally, totalling , we get the answer $1+15+20+18 =54$   
