Question

The value of ${}^{95}{C}_4+\sum _{j=1}^5{}^{100-j}{C}_3$ is

• A. ${}^{95}{C}_5$
• B. ${}^{100}{C}_4$
• C. ${}^{99}{C}_4$
• D. ${}^{100}{C}_5$

Answer 1

Answer: (B)

${}^{100}{C}_4$

${}^{95}{C}_4+\sum _{j=1}^5{}^{100-j}{C}_3$
${=}^{95}{C}_4{+}^{99}{C}_3{+}^{98}{C}_3{+}^{97}{C}_3{+}^{96}{C}_3{+}^{95}{C}_3$
$={(}^{95}{C}_3{+}^{95}{C}_4){+}^{96}{C}_3{+}^{97}{C}_3{+}^{98}{C}_3{+}^{99}{C}_3$
$={(}^{96}{C}_4{+}^{96}{C}_3){+}^{97}{C}_3{+}^{98}{C}_3{+}^{99}{C}_3$ (${\because }^n{C}_r{+}^n{C}_{r-1}={}^{n+1}{C}_r$)
$={(}^{97}{C}_4{+}^{97}{C}_3){+}^{98}{C}_3{+}^{99}{C}_3$ (${\because }^n{C}_r{+}^n{C}_{r-1}={}^{n+1}{C}_r$)
$={(}^{98}{C}_4{+}^{98}{C}_3){+}^{99}{C}_3$ (${\because }^n{C}_r{+}^n{C}_{r-1}={}^{n+1}{C}_r$)
${=}^{99}{C}_4{+}^{99}{C}_3{=}^{100}{C}_4$ (${\because }^n{C}_r{+}^n{C}_{r-1}={}^{n+1}{C}_r$)