Calculate Probability of arrangements

Question

I am trying to answer following question

Mr. Flowers plants $10$ rose bushes in a row. Eight of the bushes are white and two are red, and he plants them in random order. What is the probability that he will consecutively plant seven or more white bushes?
(It's based on the textbook "PROBABILITY AND MATHEMATICAL STATISTICS" by Professor Prasanna Sahoo)

The answer at the back of the textbook is $1/5$

Based on my calculation:

Sample Space $ = \frac{10!}{8!2!} = 45$

Number of ways to get 7 consecutive white $= \frac{4!}{2!*2!} = 6$

Number of ways to get 8 consecutive white $= \frac{3!}{2!} = 3$

However, i think there is duplication on both cases above, they are

WWWWWWWWRR

RRWWWWWWWW

So, I just use $6 + 3 - 2 = 7$, which makes the answer $7/45$

I am not sure why the textbook gave answer $1/5$.

Would someone please point to me where my mistake is?

Thank you so much!

[EDIT - 2020-08-28]: thank you all for the tips. I finally get it. I also draw all the possible outcomes.

Answer 1

The two cases are

exactly $8$ consecutive whites, which include $3$ permutations $$W^8R^2\\ R^2W^8 \\ RW^8R$$

or

exactly $7$ consecutive whites, so that there should be at least one red between $W$ and $W^7$, which include $$\underline{W^7RW}R \\R\underline{W^7RW} \\ \underline{W^7R^2W}$$ and similar three triples with the positions of $W^7$ and $W$ interchanged, making $6$ more permutations not counted before, (since there's no $W^8$ here) giving a total of $9$.

Answer 2

I don't quite follow where you see duplication, but there are $9$ distinct possibilities. If there are $8$ W's in a row, then there must be an R on either side, or two R's to the left, or two R's to the right, making $3$ cases, as you said.

Now suppose there are exactly $7$ W's in a row. There may be an R on either side, with the eighth W coming first or last, giving $2$ cases.

Otherwise, the two R's and the eighth W all come on the left or all come on the right. In the former case, the only possible arrangements are WRR and RWR. In the latter case, we must have RRW or RWR, giving another $4$ cases.