Question
The value of $$^{40}\,{C_{31}} + \sum\limits_{j = 0}^{10} {^{40 + j}{C_{10 + j}}} $$ is equal to
A.
$$^{51}{C_{20}}$$
B.
$$2{ \cdot ^{50}}\,{C_{20}}$$
C.
$$2{ \cdot ^{45}}\,{C_{15}}$$
D.
None of these
Answer :
$$^{51}{C_{20}}$$
Solution :
Value $$ = \left( {^{40}{C_9} + {\,^{40}}{C_{10}}} \right) + {\,^{41}}{C_{11}} + ..... + {\,^{50}}{C_{20}}$$
$$\eqalign{
& = \left( {^{41}{C_{10}} + {\,^{41}}{C_{11}}} \right) + {\,^{42}}{C_{12}} + ..... + {\,^{50}}{C_{20}} \cr
& = ..... = {\,^{50}}{C_{19}} + {\,^{50}}{C_{20}} = {\,^{51}}{C_{20}}. \cr} $$