How many arrangements of $A, B, C, D, E, F, G, H$ are there such that $A$
and B are between C and D ?
Attempt : I am trying to solve this problem using simple counting process like first place can be occupy by 6 persons second occupied by 6 and so forth I tried to manipulate remaining one. But it seems confusing and I am missing some terms is what I feel . Please help me to solve this with effective approach.
If you didn't have the constraint about $A,B$ being between $C,D$, then you'd have $8!=40,320$ arrangements. Now consider the positions of $A,B,C,D$ in each of those arrangements. In each arrangement, exactly one of the six pairs $AB$, $AC$, $AD$, $BC$, $BD$, and $CD$ will lie between the other two of $A,B,C,D$. So you can classify the $40,320$ arrangements into $6$ classes according to which pair lies between the others. Each of those $6$ classes has, by symmetry, the same number of elements, $40,320\,/\,6$. And your problem asks for the number of arrangements in one of the classes. So the answer is $40,320\,/\,6=6720$.