Permutation and Combiantion(Selection with condition) A town council plans to plant 12 trees along the center of the main road. The council buys the trees
from a garden center which has 4 different hibiscus trees, 9 different jacaranda trees, and 2 different
oleander trees for sale.
(i) How many different selections of 12 trees can be made if there must be at least 2 of each type of
tree?
I tried this question in two ways
(1)  $(4C2\times9C8\times2C2)+(4C3\times 9C7 \times 2C2)+(4C4\times9C6\times2C2)= 282$
(2) First of all, I am selecting two trees from each of the types. Remaining I need to select 6 more to complete 12 trees of selection. Total there is 4+9+2=15 trees are there. So, after selecting two from each type the remaining is 9(15-6=9). From this 9 I am selecting 9. So, the number of ways of selection is $4C2\times9C2\times2C2\times9C6=18144$
I found out that the answer to this question is the first one(282). But I dont understand what is the problem with second way and what exactly it is?
Sincere thanks in advance
 A: Let the trees be denoted by $H,J,O.$ You have no other choice left other than taking $2$ $O$'s. Now there are $10$ trees remaining to select. Which can be taken in three ways.
$(1)$ $2\ H, 8\ J.$
$(2)$ $3\ H, 7\ J.$
$(3)$ $4\ H, 6\ J.$
For $(1)$ you have $\binom {4} {2} \times \binom {9} {8} \times \binom {2} {2} = 54$ arrangements, for $(2)$ you have $\binom {4} {3} \times \binom {9} {7} \times \binom {2} {2} = 144$ arrangements, for $(3)$ you have $\binom {4} {4} \times \binom {9} {6} \times \binom {2} {2} = 84$ arrangements.
So in total there are $54+144+84 = 282$ arrangements.
The second method is wrong because there are repetitions. Let the $4$ $H$'s be denoted by $H_1, \cdots ,H_4,$ $9$ $J$'s be denoted by $J_1,\cdots,J_9,$ $2$ $O$'s be denoted by $O_1,O_2.$
Suppose in the first attempt you take $H_1,H_2,J_1,J_2,O_1,O_2$ and suppose in the second attempt you take $H_3,H_4,J_3,J_4,J_5,J_6.$ Now consider another case where in the first attempt you take $H_3,H_4,J_1,J_2,O_1,O_2$ and in the second attempt you take $H_1,H_2,J_3,J_4,J_5,J_6.$ 
Is there any significant difference between these two cases?
