This question has been answered before, but I want an elaboration as to why:
If there are 10 people including A and B, how many arrangements in a line(I guess this is permutation since order does matter), how many of these arrangements are A and B together?
The answer is $2!9!$, but why? I don't understand the intuition because intuitively, it seems like it should be $2!8!$:
This is a word formation problem, we need two letters (A and B) to be side by side, the possible number of ways to arrange two letters side by side is $2!$, leaving use 8 other letters to choose from to create the remaining 8 spaces which is $8!$, so $2!8!$
My book says that we treat the pair as a single object, so there are $2!$ to arrange this object, and then $(n - 1)! = 9!$ to arrange the remaining people. But doesn't this include combinations where A and B are not together since $9!$ considers possibilities of where A is not seated next to B?
Why is my reasoning incorrect and why is the book reasoning correct (or why is $2!8!$ wrong and $2!9!$ correct)?
$\endgroup$ 23 Answers
$\begingroup$Another point of view: we have to place $8$ persons, some of them on the left of the pair $AB$, the rest on the right.
Let's say $k$ persons are on the left. We have $\dbinom 8k$ ways to choose them, and have $k!$ permutations of these persons. Similarly, there are $(8-k)!$ permutations of the persons on the right, so finally there are $\;2!\dbinom8k k!\,(8-k)!=2!\,8!$ arrangements with $k$ persons on the left.
This makes in all: $$\sum_{k=0}^82!\,8!=9\cdot2!\,8!=2!\,9!$$
$\endgroup$ $\begingroup$Intuition is very simple here, we look at A and B like it is one person, then we have to arrange 9 person, it is $9!$. One AB and $8$ others.
And we can say first A than B, or first B than A, so:
$$2!9!$$
$\endgroup$ $\begingroup$This is because 9 times 8! Is 9!
$\endgroup$ 9
