Question

${}^{14}{C}_4+\sum _{j=1}^4{}^{(18-j)}{C}_3=$

• A. ${}^{14}{C}_5$
• B. ${}^{18}{C}_5$
• C. ${}^{18}{C}_4$
• D. ${}^{19}{C}_4$

Answer 1

The correct option is C ${}^{18}{C}_4$

We know, ${}^n{C}_r{+}^n{C}_{r+1}{=}^{n+1}{C}_{r+1}$

${}^{14}{C}_4+{\sum }_{j=1}^4{}^{(18-j)}{C}_3{=}^{14}{C}_4{+}^{17}{C}_3{+}^{16}{C}_3{+}^{15}{C}_3{+}^{14}{C}_3$

${=}^{17}{C}_3{+}^{16}{C}_3{+}^{15}{C}_3{+}^{15}{C}_4$

${=}^{18}{C}_4$