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) $

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