Question
The value of $$^{50}{C_4} + \sum\limits_{r = 1}^6 {^{56 - r}{C_3}} $$ is
A.
$$^{55}{C_4}$$
B.
$$^{55}{C_3}$$
C.
$$^{56}{C_3}$$
D.
$$^{56}{C_4}$$
Answer :
$$^{56}{C_4}$$
Solution :
$$\eqalign{
& ^{50}{C_4} + \sum\limits_{r = 1}^6 {^{56 - r}{C_3}} \cr
& \Rightarrow \,{\,^{50}}{C_4} + \left[ {^{55}{C_3} + {\,^{54}}{C_3} + {\,^{53}}{C_3} + {\,^{52}}{C_3} + {\,^{51}}{C_3} + {\,^{50}}{C_3}} \right] \cr
& {\text{We know }}\left[ {^n{C_r} + {\,^n}{C_{r - 1}} = {\,^{n + 1}}{C_r}} \right] \cr
& \Rightarrow \,\,\left( {^{50}{C_4} + {\,^{50}}{C_3}} \right) + {\,^{51}}{C_3} + {\,^{52}}{C_3} + {\,^{53}}{C_3} + {\,^{54}}{C_3} + {\,^{55}}{C_3} \cr
& \Rightarrow \,\,\left( {^{51}{C_4} + {\,^{51}}{C_3}} \right) + {\,^{52}}{C_3} + {\,^{53}}{C_3} + {\,^{54}}{C_3} + {\,^{55}}{C_3} \cr} $$
Proceeding in the same way, we get
$$ \Rightarrow \,{\,^{55}}{C_4} + {\,^{55}}{C_3} = {\,^{56}}{C_4}.$$