confusion about permutation $7$ 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.
 A: Number of unrestricted permutations $=\frac{10!}{7!\cdot 3!}=120$ 
Number of permutations where at least 2 black balls are together $=\frac{9!}{7!}=72$ 
Here we have considered $2$ black balls to be a single unit since they are always together and then arranged the $9$ balls.
Now we subtract the cases where the $3$ black balls are together i.e. $\frac{8!}{7!}=8$ in the same procedure as above since we have calculated the $3$ balls being together in the $72$ cases above.
So the probability $=1-\large\frac{72-8}{120}=\color{blue}{\frac{7}{15}}$ 
See if this helps..
A: Method 1: Split into two cases
not ending in $\text{B}$: $$\underbrace{\vphantom{}\text{BW}\mid\text{BW}\mid\text{BW}\mid\text{W}\mid\text{W}\mid\text{W}\mid\text{W}\vphantom{}}_{\large\binom{7}{3}}$$
ending in $\text{B}$: $$\underbrace{\vphantom{}\text{BW}\mid\text{BW}\mid\text{W}\mid\text{W}\mid\text{W}\mid\text{W}\mid\text{W}\vphantom{}}_{\large\binom{7}{2}}\ (\text{B})$$

Method 2: Add a throw-away $\bf{W}$ to the right
$$\underbrace{\vphantom{}\text{BW}\mid\text{BW}\mid\text{BW}\mid\text{W}\mid\text{W}\mid\text{W}\mid\text{W}\mid\text{W}\vphantom{}}_{\large\binom{8}{3}}$$

No restrictions
$$\underbrace{\vphantom{}\text{B}\mid\text{B}\mid\text{B}\mid\text{W}\mid\text{W}\mid\text{W}\mid\text{W}\mid\text{W}\mid\text{W}\mid\text{W}\vphantom{}}_{\large\binom{10}{3}}$$
Therefore, the probability is $$\frac{\binom{7}{3}+\binom{7}{2}}{\binom{10}{3}}=\frac{\binom{8}{3}}{\binom{10}{3}}=\frac{7}{15}$$
A: 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 $Pr = \dfrac{\binom83}{\binom{10}3} = \dfrac7{15}$ 
A: Use inclusion/exclusion principle in order to count the number of such permutations:


*

*Include the total number of permutations: $10!$

*Exclude the number of permutations with at least $\color\red2$ black balls together: $\binom{3}{\color\red2}\cdot\color\red2!\cdot9!$

*Include the number of permutations with at least $\color\red3$ black balls together: $\binom{3}{\color\red3}\cdot\color\red3!\cdot8!$


Then, in order to compute the probability, divide it by the total number of permutations:
$$\frac{10!-\binom{3}{2}\cdot2!\cdot9!+\binom{3}{3}\cdot3!\cdot8!}{10!}=\frac{7}{15}$$

Please note that in order to calculate probability, we are considering the balls as distinct rather than identical (hence the use of the term permutations).
