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\sum_{\mathrm{k}=0}^6{ }^{51} \mathrm{C}_3 is equal to

Option: 1

{ }^{51} \mathrm{C}_4-{ }^{45} \mathrm{C}_4


Option: 2

{ }^{52} \mathrm{C}_3-{ }^{45} \mathrm{C}_3


Option: 3

{ }^{52} \mathrm{C}_4-{ }^{45} \mathrm{C}_4


Option: 4

{ }^{51} \mathrm{C}_3-{ }^{45} \mathrm{C}_3


Answers (1)

best_answer

{ }^{51} \mathrm{C}_3+{ }^{50} \mathrm{C}_3+{ }^{49} \mathrm{C}_3+{ }^{48} \mathrm{C}_3+{ }^{47} \mathrm{C}_3+{ }^{46} \mathrm{C}_3+{ }^{45} \mathrm{C}_3

add and subtract { }^{45} \mathrm{C}_4

\begin{aligned} & \left({ }^{45} \mathrm{C}_4+{ }^{45} \mathrm{C}_3\right)+{ }^{46} \mathrm{C}_3+{ }^{47} \mathrm{C}_3+{ }^{48} \mathrm{C}_3+{ }^{49} \mathrm{C}_3+{ }^{50} \mathrm{C}_3+{ }^{51} \mathrm{C}_3-{ }^{45} \mathrm{C}_4 \quad\left({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}={ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}}\right) \\ & { }^{52} \mathrm{C}_4-{ }^{45} \mathrm{C}_4 \\ & \Rightarrow[\mathrm{C}] \end{aligned}

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Anam Khan

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