MCQ
$\sum \limits_{ k =0}^6{ }^{51- k } C _3$ is equal to
  • A
    ${ }^{51} C _4-{ }^{45} C _4$
  • B
    ${ }^{51} C _3-{ }^{45} C _3$
  • ${ }^{52} C _4-{ }^{45} C _4$
  • D
    ${ }^{52} C _3-{ }^{45} C _3$

Answer

Correct option: C.
${ }^{52} C _4-{ }^{45} C _4$
c
$\sum \limits_{ k =0}^6{ }^{51- k } C _3$

$={ }^{51} C _3+{ }^{50} C _3+{ }^{49} C _3+\ldots+{ }^{45} C _3$

$={ }^{45} C _3+{ }^{46} C _3+\ldots \ldots+{ }^{51} C _3$

$={ }^{45} C _4+{ }^{45} C _3+{ }^{46} C _3+\ldots \ldots+{ }^{51} C _3-{ }^{45} C _4$

$\left.={ }^n C _{ r }+{ }^n C _{ r -1}={ }^{ n +1} C _{ r }\right)$

$={ }^{52} C _4-{ }^{45} C _4$

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