Seating four girls and two boys in a row such that the boys do not sit together If $2$ boys are never to sit together and $4$ girls and $2$ boys are to sit in linear line.? Then total number of such arrangements is:
My solution:
The total number of linear arrangements is $6!$ and the number of arrangements when $2$ boys are to sit together is $5!$ so the answer should be $6!-5!=600$. Am I right here?
 A: Answer is $6!-2.5!$ as two boy can sit together in a two different way.
A: Consider 2 boys as a single unit .. then possible ways of arranging boys & 4 girls = arrangements such  that 2 boys sit together  => 5! => 120 
Now ,since 2 boys are never to sit together ,
the number of required arrangements  => 6! -(2*120) => 720-240 => 480
A: Imagine you have four "girls-only" seats :
$$ \sqcup \quad \sqcup \quad \sqcup \quad \sqcup$$
and you want to insert two "boys-only" seats, but not side-by-side, that is among the points in :
$$ \cdot \sqcup \cdot \sqcup \cdot \sqcup \cdot \sqcup \cdot$$
Then you need to choose $2$ spots among the $5$ dots : there is $\begin{pmatrix}5\\2\end{pmatrix}$ such possibilities.
Now choose one of those possibilities : to seat the four girls on the "girls-only" seats, you have $4!$ ways of doing it ; for each of those ways, you need to put the boys in the "boys-only" seats, for which there is $2$ ways. So for each seats order, you have $2 \times 4!$ ways of putting the girls and boys in the seats.
At the end, you have $2 \times 4! \times\begin{pmatrix}5\\2\end{pmatrix} = 4! \times 20$ possible arrangements.
A: Considering only 'boy' $\textbf{b}$ and 'girl' $\textbf{g}$ we only have
$$
\frac{6!}{2!4!} = 15
$$
permutations, they are given by
$$
\textbf{bbgggg} - \textbf{bgbggg} - \textbf{bggbgg} - \textbf{bgggbg} - \textbf{bggggb}\\
\textbf{gbbggg} - \textbf{gbgbgg} - \textbf{gbggbg} - \textbf{gbgggb}\\
\textbf{ggbbgg} - \textbf{ggbgbg} - \textbf{ggbggb}\\
\textbf{gggbbg} - \textbf{gggbgb}\\
\textbf{ggggbb}\\
$$
The condition that two boys never sit together means to exclude permutation containing 
$$
\textbf{bb}
$$
That are $5$ permutations, so we obtain
$$
16 - 5 = 10
$$
permutations, they are given by
$$
\textbf{bgbggg} - \textbf{bggbgg} - \textbf{bgggbg} - \textbf{bggggb}\\
\textbf{gbgbgg} - \textbf{gbggbg} - \textbf{gbgggb}\\
\textbf{ggbgbg} - \textbf{ggbggb}\\
\textbf{gggbgb}\\
$$
The general case is given by
$$
\frac{\big(b+g\big)!}{b!g!} - g - 1
$$

If we label the 'boys' and 'girls' we need to multiply by
$$
b!g!
$$
so we get
$$
\big(b+g\big)! - b!\big(g+1\big)!
$$
As in this case we get
$$
6! - 2! 5! = 480
$$
