In how many ways can a photographer arrange $8$ people in a row from a family of $10$
people, if
(a) the bride and groom are in the photo. This would be $9! = 9\times8\times7\times6\times5\times4\times3\times2\times1=362880$, correct?
(b) the bride and groom are next to each other in the photo
(c) either the bride or the groom is in the photo, not both
I'm not sure how to start $b$ or $c$. I know $b$ is more restricting than $a$ but I don't know how to represent it
Best Answer
b) Treat the bride and groom as one block. Then we need to choose $6$ out of the remaining $8$ people. Then we can arrange the $7$ blocks as $7!$ but we need to multiply by $2$ since we can have GB or BG.
c) Count the complement. A represents all possibilities and then count when BOTH are in the photo (which is just 6 choose 8). Subtract.