Question

The council buys 4 hibiscus trees, 6 jacaranda trees and 2 oleander trees.How many different arrangements of these 12 trees can be made if no hibiscus tree is next to
another hibiscus tree?

Answer 1

first arrange the ja-something tree and 2ol-something tree.
you will get, 8!/(6!2!) = 28

Answer 2

then choose some combination of these two types of 8 trees ...
... then pick 6 number from 0, 1, 2 .... 8.
suppose your number is 0, 1, 2, 3, 4, 5
then you insert hib-something tree after each of these trees.
you can do this in 8C6 = 28.

I think the answer might be 28x28

Answer 3

It's done something like this:
\[8! \times 9P4\]

Answer 4

what does that value give?

Answer 5

hold on a sec :| i made some error in the question.

Answer 6

don't choose six trees, there are only 4 hib-something trees.

Answer 7

so choose only 4 trees.

Answer 8

and answer 8!x9P4 is correct if and only if ... each tree is distinguishable from each other.

Answer 9

also I wrote ... 0, 1, 2 .... 8 = nine choices
but I took only 8 ... damn!! stupid me!!

Answer 10

Oh okay. Thank you.

Answer 11

did you get the answer?? I mean did you need any explanations?

Answer 12

Yes, got it. Like there are 81 ways the other trees can be arranged. And since there would be 9 gaps which could be filled and only 4 trees, so 9P4. Am I right?

Answer 13

* 8!

Answer 14

yes yes ... this is how I did stars and bars problem.