If there are $11$ presents which include present $A, B$, and $C$. How many ways are there to arrange $11$ presents in a line such that present $A$ and $B$ are always ahead of (not necessarily immediately ahead) of present C.
My approach: I know that in questions like this, we can say by symmetry there are as many ways to $X$ comes before $Y$ as $Y$ comes before $X$ and in that case, the number of ways is (total number of presents)!/2. Similarly, here we can bundle present $A$ and $B$ together and call it present $Z$. There are $10!$ (we count $A$ and $B$ as a single bundle) ways to arrange them all. And $(10!)/2$ ways where bundle $Z$ comes before present $C$. We then multiply $(10!/2) * 2$ to account for present $A$ being in front of $B$ and vice versa in the bundle $Z$.
This is wrong. Presents $A $ and $B $ may not be together but still ahead of present $C$. For example, this one- $$***A*B****C$$ Therefore, we cannot tie them for arranging.
The correct approach is that we know that there are $11!$ total possible arrangements. Out of them, by symmetry, each of the following arrangement among $A,B$ and $ C$ is equally likely- $$ABC$$ $$BAC$$ $$ACB$$ $$BCA$$ $$CAB$$ $$CBA$$ Out of these, first two are what we want and others are unfavourable. So, total number of arrangements are $$11!×\frac{2}{6}=\frac{11!}{3}$$
which is the required answer.
Hope it helps:)