$\binom{47}{4} + \sum_{r=1}^5 \binom{52-r}{3} = \dots$

  • A
    $\binom{47}{6}$
  • B
    $\binom{52}{5}$
  • C
    $\binom{52}{4}$
  • D
    $\binom{52}{3}$

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