5 boys and 6 girls are arranged in all possible ways. Let X denote the number of linear arrangements in…
5 boys and 6 girls are arranged in all possible ways. Let X denote the number of linear arrangements in which no two boys sit together and $Y$ denote the number of linear arrangements in which no two girls sit together. If $Z$ denote the number of ways of arranging all of them around a circular table such that no two boys sit together, then $X: Y: Z=$
$1: 1: 21$
$21: 1: 1$
$7: 5: 5$
$4: 3: 3$
Solution
When no two boys sit together,
$* \mathrm{G} * \mathrm{G} * \mathrm{G} * \mathrm{G} * \mathrm{G} * \mathrm{G} *$
$\therefore X=$ Number of ways of arrangement in which no two boys sit together $={ }^7 C_5 \times 5!\times 6!$
When no two girls sit together
${ }^* \mathrm{~B} * \mathrm{~B} * \mathrm{~B} * \mathrm{~B} * \mathrm{~B} *$
$\therefore Y=$ Number of ways of arrangement in which no two girls sit together $={ }^6 C_6 \times 5!\times 6$ !
If $Z=$ Number of ways of arranging all of them around a circular toable such that no two boys sit together
$\begin{aligned}
& ={ }^6 C_5 \times 5!\times 5! \\
& \therefore X: Y: Z={ }^7 C_5 \times 5!\times 6!\cdot{ }^6 C_6 \times 5!\times 6!\cdot{ }^6 C_5 \times 5!\times 5! \\
& \quad=21: 1: 1
\end{aligned}$