<s>[INST] Seating arrangements with no 3 objects together. 
Suppose that five $1$'s and six $0$'s need to be arranged in such a way that no three $0$'s are consecutive. How many different arrangements are possible? 

This is a variation on a problem where some number of boys and girls need to be seated such that no two boys sit next to each other. What are the solution strategies if the number of boys is increased to $3$, $4$, etc?
Thank you.
 [/INST] user84413 has provided an elegant approach to solving this problem.  
Here is another approach:
If we place five ones in a row, we create six spaces, four between successive ones and two at the ends, which we wish to fill with six zeros.  If we let $x_k$ denote the number of zeros placed in the $k$th space from the left, then the number of ways we can arrange five ones and six zeros so that at most two zeros are consecutive is the number of nonnegative integer solutions of the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 6 \tag{1}$$
subject to the restrictions that $x_k \leq 2$ for $1 \leq k \leq 6$.
When there were no restrictions, then the number of solutions of equation 1 in the nonnegative integers is equal to the number of ways we can place five addition signs in a row of six ones, which is $\binom{6 + 5}{5} = \binom{11}{5}$.  However, we must exclude those cases in which three or more zeros are consecutive.  
Suppose that $x_1 \geq 3$.  Let $y_1 = x_1 - 3$.  Then $y_1$ is a nonnegative integer and
\begin{align*}
x_1 + x_2 + x_3 + x_4 + x_5 + x_6 & = 6\\
y_1 + 3 + x_2 + x_3 + x_4 + x_5 + x_6 & = 6\\
y_1 + x_2 + x_3 + x_4 + x_5 + x_6 & = 3 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers, with $\binom{3 + 5}{5} = \binom{8}{5}$ solutions.  Since there are six ways one of the variables could exceed $2$, we must subtract $\binom{6}{1}\binom{8}{5}$ from $\binom{11}{5}$.  However, doing so subtracts those cases in which there are two distinct blocks of three zeros twice. There are $\binom{6}{2}$ ways to fill two of the six spaces with three zeros.  Therefore, by the Inclusion-Exclusion Principle, the number of bit strings consisting of five ones and six zeros in which at most two zeros are consecutive is 
$$\binom{11}{5} - \binom{6}{1}\binom{8}{5} + \binom{6}{2}$$
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