Question

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

• A. ${ }^{51} C _3-{ }^{45} C _3$
• B. ${ }^{52} C _4-{ }^{45} C _4$
• C. ${ }^{52} C _3-{ }^{45} C _3$
• D. ${ }^{51} C _4-{ }^{45} C _4$

Answer 1

Answer: (B)

${ }^{52} C _4-{ }^{45} C _4$

Answer 2

The correct answer is (B) : ${ }^{52} C _4-{ }^{45} C _4$
k=0∑6​51−kC3​
=51C3​+50C3​+49C3​+…+45C3​
=45C3​+46C3​+…..+51C3​
=45C4​+45C3​+46C3​+…..+51C3​−45C4​
(nCr​+nCr−1​=n+1Cr​)
=52C4​−45C4​