JEE MAIN - Mathematics (2023 - 25th January Evening Shift - No. 16)
$$\sum\limits_{k = 0}^6 {{}^{51 - k}{C_3}} $$ is equal to :
$$\mathrm{{}^{51}{C_4} - {}^{45}{C_4}}$$
$$\mathrm{{}^{51}{C_3} - {}^{45}{C_3}}$$
$$\mathrm{{}^{52}{C_3} - {}^{45}{C_3}}$$
$$\mathrm{{}^{52}{C_4} - {}^{45}{C_4}}$$
Explanation
$$
\begin{aligned}
& \sum_{\mathrm{k}=0}^6{ }^{51-\mathrm{k}} \mathrm{C}_3 \\\\
& ={ }^{51} \mathrm{C}_3+{ }^{50} \mathrm{C}_3+{ }^{49} \mathrm{C}_3+\ldots+{ }^{45} \mathrm{C}_3 \\\\
& ={ }^{45} \mathrm{C}_3+{ }^{46} \mathrm{C}_3+\ldots \ldots+{ }^{51} \mathrm{C}_3 \\\\
& ={ }^{45} \mathrm{C}_4+{ }^{45} \mathrm{C}_3+{ }^{46} \mathrm{C}_3+\ldots . .+{ }^{51} \mathrm{C}_3-{ }^{45} \mathrm{C}_4 \\\\
& \left({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}={ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}}\right) \\\\
& ={ }^{52} \mathrm{C}_4-{ }^{45} \mathrm{C}_4
\end{aligned}
$$
Comments (0)
