(D) The number of ways to arrange $m$ boys and $m$ girls in a row such that no two boys sit together is $x = (m+1)! m!$. The number of ways to arrange

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(D) The number of ways to arrange $m$ boys and $m$ girls in a row such that no two boys sit together is $x = (m+1)! m!$. The number of ways to arrange $m$ boys and $m$ girls in a row such that they sit alternately is $y = m! \times m! \times 2$. The number of ways to arrange $m$ boys and $m$ girls around a circular table such that they sit alternately is $z = (m-1)! m!$. Thus,the ratio is: $x: y: z = (m+1)! m! : 2(m! m!) : (m-1)! m!$ Dividing by $(m-1)! m!$,we get: $x: y: z = (m+1)m : 2m : 1$.

Class 11 Mathematics Permutation and Combination Mix Examples-Permutation and Combination

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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 $z$ denote 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 $x: y: z=$

TS EAMCET 2019 All TS EAMCET

A
$m+1: m: m-1$
B
$3: 2: 1$
C
$m-1: m: 2$
D
$(m+1)m: 2m: 1$

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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 $z$ denote 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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