Question

The value of ${}^{50}{C}_4+{\sum }_{r=1}^6{}^{56-r}{C}_3$ is

• A. ${}^{55}{C}_4$
• B. ${}^{55}{C}_3$
• C. ${}^{56}{C}_3$
• D. ${}^{56}{C}_4$

Answer 1

The correct option is D ${}^{56}{C}_4$

${}^{50}{C}_4+{\sum }_{r=1}^6{}^{56-r}{C}_3$ ${\Rightarrow }^{50}{C}_4+{[}^{55}{C}_3{+}^{54}{C}_3{+}^{53}{C}_3{+}^{52}{C}_3{+}^{51}{C}_3{+}^{50}{C}_3]$

$={(}^{50}{C}_4{+}^{50}{C}_3){+}^{51}{C}_3{+}^{52}{C}_3{+}^{53}{C}_3{+}^{54}{C}_3{+}^{55}{C}_3$ ........................ As ${(}^n{C}_r{+}^n{C}_{r-1}{=}^{n+1}{C}_r)$

$={(}^{51}{C}_4{+}^{51}{C}_3){+}^{52}{C}_3{+}^{53}{C}_3{+}^{54}{C}_3{+}^{55}{C}_3$

$={}^{52}{C}_4{+}^{52}{C}_3{+}^{53}{C}_3{+}^{54}{C}_3{+}^{55}{C}_3$

$={}^{53}{C}_4{+}^{53}{C}_3{+}^{54}{C}_3{+}^{55}{C}_3$

$={}^{54}{C}_4{+}^{54}{C}_3{+}^{55}{C}_3$

${\Rightarrow }^{55}{C}_4{+}^{55}{C}_3{=}^{56}{C}_4$