Binomial Theorem Question 7
Question 7 - 25 January - Shift 2
$\sum _{k=0}^{6}{ }^{51-k} C_3$ is equal to
(1) ${ }^{51} C_4-{ }^{45} C_4$
(2) ${ }^{51} C_3-{ }^{45} C_3$
(3) ${ }^{52} C_4-{ }^{45} C_4$
(4) ${ }^{52} C_3-{ }^{45} C_3$
Show Answer
Answer: (3)
Solution:
Formula: Properties of Binomial Coefficients (iv)
$\sum _{k=0}^{6}{ }^{51-k} C_3$
$={ }^{51} C_3+{ }^{50} C_3+{ }^{49} C_3+\ldots+{ }^{45} C_3$
$={ }^{45} C_3+{ }^{46} C_3+\ldots \ldots+{ }^{51} C_3$
$={ }^{45} C_4+{ }^{45} C_3+{ }^{46} C_3+\ldots \ldots+{ }^{51} C_3-{ }^{45} C_4$
$ ({ }^{n} C_r+{ }^{n} C _{r-1}={ }^{n+1} C_r) $
$={ }^{52} C_4-{ }^{45} C_4$
