The corners of regular tetrahedrons are numbered $1, 2, 3, 4$. Three tetrahedrons are tossed. The probability that the sum of the upward corners will be $5$ is

  • A
    $\frac{5}{24}$
  • B
    $\frac{5}{64}$
  • C
    $\frac{3}{32}$
  • D
    $\frac{3}{16}$

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