TITLE: confusion about permutation
QUESTION [11 upvotes]: $7$ white identical balls and $3$ black identical balls are randomly placed in a row. The probability that no two black balls are together is ? 
I am getting it as $ \frac{1}{3}$ while the answer in my book is $\frac{7}{15}$. Total ways are $\frac{10!}{7!.3!}=120$ now i considered three consecutive balls as one so $\frac{(1+7)!}{7!}=8$,then two balls as consecutive which is $\frac{(1+8)!}{7!}=72$ so probability is $\frac{120-8-72}{120}=1/3$
What am I missing on? Any help.

REPLY [7 votes]: Another way
Consider a string of $7$ white balls. There are $8$ places between balls (including ends) where black balls may be inserted w/o being adjacent, against $\binom{10}3$ unrestricted arrangements
$\uparrow\huge\circ$$\uparrow\huge\circ$$\uparrow\huge\circ$$\uparrow\huge\circ$$\uparrow\huge\circ$$\uparrow\huge\circ$$\uparrow\huge \circ$$\uparrow$
thus $Pr = \dfrac{\binom83}{\binom{10}3} = \dfrac7{15}$