Seating four girls and two boys in a row such that the boys do not sit together

Question

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?

Answer 1

Answer is $6!-2.5!$ as two boy can sit together in a two different way.

Answer 2

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 => $$6! -(2\times120)$$ $$=720-240$$ $$=480$$

Answer 3

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.

Answer 4

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

$$ 6! - 2! 5! = 480 $$