Nine stuffed animals, all different, are placed along a babies toy shelf. How many possible
arrangements are there if BERT and ERNIE must not be together.
Thanks in advance
Let’s start with the number of possible arrangements. This is equal to 9!, or 362,880.
Now let’s write out the animals as numbers 1-9.
1 2 3 4 5 6 7 8 9
Bert and Ernie can be next to each other at 12, 23, 34, 45, 56, 67, 78, or 89, which eliminates 8 arrangements. However, they can either be placed with Bert first and Ernie second or Ernie first and Bert second. This means that at each of the 8 positions that the two are together, they can be arranged 2 ways. This means that the number of arrangements meeting this requirement is 362,880 - (8 x 2) = 362,864.
Let me know if this works/helps.
@Hermit9 you forgot to multiply by 7! to account for placing the other seven.
ways = 9! - 827! = 282240
@MITer94 Yep, thank you!
@Hermit9 also for these kinds of questions, it may be easier to treat Bert and Ernie as a single stuffed animal so that you don’t have to count the positions.
ok thanks again @MITer94