A farmer has 30 plants: 12 Roses, 12 Jasmine and 6 Lotus, How many possibilities can he plant them in a row A farmer has 30 plants: 12 Roses, 12 Jasmine and 6 Lotus, How many possibilities can he plant them in a row if 2 lotuses can't grow next to each other.
I tried looking at the question in this way:
Say i wanna make a word with 30 letters. I'll mark Roses as R, Jasmine as J, and Lotus and L.
So i wanna make a word with exactly 12 R's, 12 J's and 6L's where i don't have 2 L's next to each other.
So the total number of possibilities to make up a word is: $\frac {30!}{12!12!6!}$.
And i want to reduce the possibilities where there's at least 2L's in a row, So i'll define a new 'letter' T which is LL, And get $\frac {29!}{12!12!4!1!}$.
But than by doing $\frac {30!}{12!12!6!}-\frac {29!}{12!12!4!1!}$ i get $0$.
What's is wrong with my solution?
Thanks
 A: In your second solution you count many of the possibilities a few times.
For example: once you can choose the first plane as LL and then another lotus, and you can do the opposite as well.
To solve this, you can think about the 12 Lotus planes as dividers, and than put at least one plane between each two close dividers.
A: First count the number of ways to plant the roses and jasmine in a row:
 since there are 24 plants and 12 of them are roses, there are C(24,12) choices.
Now we have 25 gaps (counting the outside gaps) in which to place the lotus plants, so there are C(25,6) ways to do this.
Therefore there are C(24,12)*C(25,6) ways to plant them in a row with no two lotus plants together.
A: I've figured the question: [Pretty similar to the answers above, But still using the idea i used when i asked the question]
Say i wanna make a word with 30 letters. I'll mark Roses as R, Jasmine as J, and Lotus and L. So i wanna make a word with exactly 12 R's, 12 J's and 6L's where i don't have 2 L's next to each other.
Firstly, i'll build a word without using the letter L: I have total of 24 letters, I want exactly 12 R's And 12 J's, So i get :$\frac {24!}{12!12!}$.
We'll look at each letter as a partition, and at every space as a cell, each cell can contain up to 1 L. I have total of 25 cells (1 At start, 23 between each of the letters, 1 At the end) and 6 L's, Then we get:$\binom {25}{6}$
Multiplying the above gives us: $\frac {24!}{12!12!}\binom {25}{6}$ and we got what is needed.
The problem with my solution was that i had 4 L's besides the group LL that made me count any permutation that contains LL a lot more than 1 time.
