Question

The value of $\stackrel{3}{\sum _{r=0}}$ ${}^8{C}_r\left({}^5{C}_{r+1}{-}^4{C}_r\right)$ equals

• A. $210$
• B. $220$
• C. $120$
• D. $110$

Answer 1

Answer: (B)

$220$

$\stackrel{3}{\sum _{r=0}}$ ${}^8{C}_r({}^5{C}_{r+1}-{}^4{C}_r)$

$=\stackrel{3}{\sum _{r=0}}$ ${}^8{C}_r\left({}^4{C}_{r+1}\right)$ ...... [Since ${}^4{C}_{r+1}{+}^4{C}_r={}^5{C}_{r+1]}$

$={}^8{C}_0{}^4{C}_1+{}^8{C}_1{}^4{C}_2+{}^8{C}_2{}^4{C}_3+{}^8{C}_3{}^4{C}_4$

$=1\cdot 4+8\cdot 6+28\cdot 4+56\cdot 1$

$=4+48+112+56$

$=220$