Six objects $O_1$ to $O_6$ are arranged one on top of the other. In how many ways can these be arranged such that $O_1$ and $O_2$ are the $2$ bottom-most objects?

  • A
    $4!$
  • B
    $4! \times 2!$
  • C
    $\frac{6!}{2!}$
  • D
    $6!$

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