Question:

$\displaystyle\sum_{k=0}^{6}{}^{51-k}C_3$ is equal to:

${}^{51}C_4-{}^{45}C_4$

${}^{51}C_3-{}^{45}C_3$

${}^{52}C_4-{}^{45}C_4$

${}^{52}C_3-{}^{45}C_3$

Step-by-Step Solution

Key Concept: $\sum_{k=0}^{6}{}^{51-k}C_3={}^{51}C_3+{}^{50}C_3+\cdots+{}^{45}C_3$. By Hockey Stick: $\sum_{i=3}^{n}{}^iC_3={}^{n+1}C_4$.

${}^{52}C_4-{}^{45}C_4$.

Correct Answer: 3