$5$ 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 $Z$ denotes 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
    $1: 1: 21$
  • B
    $21: 1: 1$
  • C
    $7: 5: 5$
  • D
    $4: 3: 3$

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