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Let $x$ denote the number of ways of arranging $m$ boys and $m$ girls in a row so that no two boys sit together. If $y$ and $z$ give the number of ways of arranging $m$ boys and $m$ girls in a row and around a circular table respectively so that boys and girls sit alternately, then $x: y: z=$
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The correct answer is:
$(m+1) m: 2 m: 1$
Number of ways of arranging $m$ boys and $m$ girls in a row such that no two boys sit together, i.e. $x=(m+1) ! m !$
Number of ways of arranging $m$ boys and $m$ girls in a row such that boys and girls sit alternately, i.e. $y=m ! \times m ! \times 2 !$
Number of ways of arranging $m$ boys and $m$ girls in a circular table such that boys and girls sit alternately, i.e. $z=(m-1) ! m !$
$$
\begin{aligned}
\therefore x: y: z= & (m+1) ! m !: m ! m ! \times 2:(m-1) ! m ! \\
& =(m+1) m: 2 m: 1
\end{aligned}
$$
Number of ways of arranging $m$ boys and $m$ girls in a row such that boys and girls sit alternately, i.e. $y=m ! \times m ! \times 2 !$
Number of ways of arranging $m$ boys and $m$ girls in a circular table such that boys and girls sit alternately, i.e. $z=(m-1) ! m !$
$$
\begin{aligned}
\therefore x: y: z= & (m+1) ! m !: m ! m ! \times 2:(m-1) ! m ! \\
& =(m+1) m: 2 m: 1
\end{aligned}
$$
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