Question

If $\frac{\sum _{r=0}^9(-1{)}^r\cdot 3^{9-r}\cdot {}^{20}{C}_r\cdot {}^{20-r}{C}_{9-r}}{2^9}={}^{20}{C}_a$ then $(a-7)$ can be

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B), (D)

$\frac{\sum _{r=0}^9(-1{)}^r\cdot 3^{9-r}\cdot \frac{20!}{r!\cdot (20-r)!}\cdot \frac{(20-r)!}{(9-r)!\cdot 11!}}{2^9}$
$=\frac{\sum _{r=0}^9(-1{)}^r\cdot \frac{3^{9-r}\cdot 9!}{r!(9-r)!}\cdot \frac{20!}{11!\cdot 9!}}{2^9}=\frac{(3-1{)}^9\cdot \frac{20!}{11!\cdot 9!}}{2^9}$
$={}^{20}{C}_9\Rightarrow a=9,11$
$\Rightarrow a-7=2,4$