Question

The sum ${}^{10}{\text{C}}_3{+}^{11}{\text{C}}_3{+}^{12}{\text{C}}_3+..........{+}^{20}{\text{C}}_3$ is equal to

• A. ${}^{21}{\text{C}}_4$
• B. ${}^{21}{\text{C}}_4+$ ${}^{10}{\text{C}}_4$
• C. ${}^{21}{\text{C}}_{17}-$ ${}^{10}{\text{C}}_6$
• D. none of these

Answer 1

The correct option is C ${}^{21}{\text{C}}_{17}-$ ${}^{10}{\text{C}}_6$

Using
${}^n{C}_r+{}^{n+1}{C}_r={}^{n+1}{C}_{r+1}$
${}^{10}{C}_3+{}^{11}{C}_3+{}^{12}{C}_3+{}^{13}{C}_3+...{}^{20}{C}_3$
$=[{}^{10}{C}_4+{:}^{10}{C}_3+{}^{11}{C}_3+{}^{12}{C}_3+{}^{13}{C}_3+...{}^{20}{C}_3]-{}^{10}{C}_4$
$=[{}^{11}{C}_4+{:}^{11}{C}_3+{}^{12}{C}_3+...{}^{20}{C}_3]-{}^{10}{C}_4$
$=[{}^{12}{C}_4+{:}^{12}{C}_3+{}^{13}{C}_3+...{}^{20}{C}_3]-{}^{10}{C}_4$
:
:
:
$={}^{20}{C}_4+{}^{20}{C}_3-{}^{10}{C}_4$
$={}^{21}{C}_4-{}^{10}{C}_4$
$={}^{21}{C}_{17}-{}^{10}{C}_6$