$\sum_{r=0}^{10} {}^{40-r} C_5$ is equal to

  • A
    ${}^{41} C_5 - {}^{30} C_5$
  • B
    ${}^{41} C_6 - {}^{30} C_6$
  • C
    ${}^{41} C_5 + {}^{30} C_5$
  • D
    ${}^{41} C_6$

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