Combinatorial problem : How many ways can you choose 3 balls from nine such that no two of them are consecutive? I tried approaching this problem by first assuming that all the balls are chosen from the odd places(10) only and then the even places only(6). And then adding them together. However, my approach seems to be wrong. 
Please help me.
 A: Use the gap method
Remove $3$ balls and place in the $7$ gaps between the $6$ balls left behind
$- \bullet-\bullet-\bullet-\bullet-\bullet-\bullet-$
This can be done in $\dbinom73 = 35$ ways
A: Number the balls $1,2,\ldots,9$ from left to right. 
You are trying to choose $3$ numbers $1 \le a<b<c \le 9$ such that $b-a > 1$ and $c-b > 1$. 
Alternatively, you can choose $3$ numbers $1 \le x<y<z \le 7$, and set $(a,b,c) = (x,y+1,z+2)$. 
Once you've verified that these two methods are equivalent, you can simply count the number of ways to choose $3$ distinct numbers from $7$.
A: There are several possible approaches.  I particularly like the method posted by true blue anil.  
Method 1:  This is another way to visualize the gap method posted by true blue anil. 
Place six blue balls in a row.  This creates seven spaces, five spaces between successive blue balls and two at the ends of the row.  Place three green balls in these seven spaces, then number the balls from left to right.  Since the green balls are placed between the blue balls or at the ends, no two of the numbers on the green balls are consecutive.  The three green balls can be placed in the seven spaces in $\binom{7}{3}$ ways.  
Method 2: This is another way of viewing the bijection described by JimmyK4542.  
Place seven balls of different colors (none of which is red) in a row.  Choose three of them.  Now insert a red ball to the immediate right of the two leftmost balls you selected from the original seven so that no two of the three balls selected are consecutive.  Number the nine balls from left to right.  The number of ways we can select three balls from the original seven is $\binom{7}{3}$.  
Method 3:  We use the Inclusion-Exclusion Principle.
We can select three of the nine balls in $\binom{9}{3}$ ways.  From these, we must exclude those selections in which at least two balls are consecutive.  
There are eight ways of choosing two consecutive balls (corresponding to the eight ways of choosing the first ball in each pair) and seven ways of selecting the third ball, giving $8 \cdot 7 = 56$ selections in which at least two balls are consecutive.
If we subtract the selections in which at least two balls are consecutive from the total number of ways of selecting three balls, we have subtracted those selections in which three consecutive balls are selected twice, once when we select the first two of the three consecutive balls and once when we select the last two of the three consecutive balls.  Thus, we must add the number of ways in which to select three consecutive balls so that we that only exclude these selections once.  We can select three consecutive balls in seven ways, corresponding to the seven ways of selecting the first ball in each trio.  
Hence, the number of ways of selecting three of the nine balls so that no two of the selected balls are consecutive is 
$$\binom{9}{3} - \binom{8}{1}\binom{7}{1} + \binom{7}{1}$$
Caveat:  Trying to extend this method to the selection of more than three non-consecutive balls can be tricky, which is why I prefer the first two methods.  
Method 4:  Let $a_k$ be the position of the $k$th ball that is selected. Let
\begin{align*}
x_1 & = a_1\\
x_2 & = a_2 - a_1\\
x_3 & = a_3 - a_2\\
x_4 & = 9 - a_3
\end{align*}
Then 
$$x_1 + x_2 + x_3 + x_4 = 9 \tag{1}$$
Since no two of the balls are consecutive, $x_2, x_3 \geq 2$.  Also, since $a_1 \geq 1$, $x_1 \geq 1$.  Let 
\begin{align*}
y_1 & = x_1 - 1\\
y_2 & = x_2 - 2\\
y_3 & = x_3 - 2\\
y_4 & = x_4
\end{align*}
Then $y_k$ is a non-negative integer for $1 \leq k \leq 4$.  Substituting $y_1 + 1$ for $x_1$, $y_2 + 2$ for $x_2$, $y_3 + 2$ for $x_3$, and $y_4$ for $x_4$ in equation 1 yields
\begin{align*}
y_1 + 1 + y_2 + 2 + y_3 + 2 + y_4 & = 9\\
y_1 + y_2 + y_3 + y_4 & = 4 \tag{2}
\end{align*}
Equation 2 is an equation in the non-negative integers.  A particular solution corresponds to the placement of three addition signs in a row of four ones.  For instance,
$$1 + + 1 + 1 1$$
corresponds to the solution $y_1 = 1$, $y_2 = 0$, $y_3 = 1$, $y_4 = 2$, while 
$$1 + 1 + 1 + 1$$
corresponds to the solution $y_1 = y_2 = y_3 = y_4 = 1$.  The number of solutions of equation 2 is the number of ways we can insert three addition signs in a row of four ones, which is $\binom{7}{3}$ since we must select which three of the seven symbols (three addition signs and four ones) will be addition signs.
A: Ill give you another way. Total ways in which no two consecutives ate selected is  total ways -(ways in which 2 are consecutive +ways in which all 3 are consecutive). So total ways are ${9\choose 3}=84$ ways where three are are consecutive are $(1,2,3)..(7,8,9)=7$ . Ways where two are consecutive. "Be careful"! See if we select $(1,2)$ and $(8,9)$ we can select any from  $6$ balls but if we have middle pairs like $(2,3)..(7,8)$ which are $6$ pairs then we can select only $5$ balls  I hope you can draw it and see the reason why its so. Hence total ways are $84-({2\choose 1}.{6\choose 1}+{6\choose 1}.{5\choose 1}+{7\choose 1})=84-(12+30+7)=84-49=35$ hope its clear.
