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

$=\left({}^{50}{C}_4{+}^{50}{C}_3\right){+}^{51}{C}_3{+}^{52}{C}_3{+}^{53}{C}_3{+}^{54}{C}_3{+}^{55}{C}_3$

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

On continuing the process, we get ${}^{56}{C}_4$