Q55 / 160 · AP EAMCET 2024 (22 May Shift 1) · Mathematics

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 $X: Y: Z=$

  • A
    $1: 1: 21$
  • Correct answer
    B
    $21: 1: 1$
  • C
    $7: 5: 5$
  • D
    $4: 3: 3$

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 $\begin{aligned} & ={ }^6 C_5 \times 5!\times 5! \\ & \therefore X: Y: Z={ }^7 C_5 \times 5!\times 6!\cdot{ }^6 C_6 \times 5!\times 6!\cdot{ }^6 C_5 \times 5!\times 5! \\ & \quad=21: 1: 1 \end{aligned}$

Hence, the correct answer is option: 'B'.

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