Question

$45C_{8} + \sum_{k = 1}^{7}{\mspace{6mu}^{52 - k}C_{7}\mspace{6mu} + \mspace{6mu}}\sum_{i = 1}^{5}{\mspace{6mu}^{57 - i}C_{50 - i}}$ is equal to

• A. $57C_{7}$
• B. $56C_{7}$
• C. $57C_{8}$
• D. None of these

Answer 1

Answer: (C)

$57C_{8}$

Answer 2

Now,

$45C_{8} + \sum_{k = 1}^{7}{52 - kC_{7} + \sum_{i = 1}^{5}{57 - iC_{50 - i}}}$

=$45C_{8} + (^{51}C_{7} +^{50}C_{7} + ... +^{45}C_{7}) + (^{56}C_{49} +^{55}C_{48} + ... +^{52}C_{45})$ = $(^{45}C_{8} +^{45}C_{7} +^{46}C_{7} + ... +^{51}C_{7}) + (^{56}C_{7} +^{55}C_{7} + ... +^{52}C_{7})$

$\because$ ( $nC_{r} =^{n}C_{n - r}$).

=$(^{45}C_{8} +^{45}C_{7}) +^{46}C_{7} +^{47}C_{7} + .... +^{56}C_{7}$= $(^{46}C_{8} +^{46}C_{7}) +^{47}C_{7} + ... +^{56}C_{7}$

= $57C_{8}$

($\because\mspace{6mu}^{n}C_{r} +^{n}C_{r - 1} =^{n - 1}C_{r}$)