Question

The value of $\sum _{r=1}^{10}r(r-1)\cdot {}^{10}{C}_r$ is

• A. $45\times 2^{10}$
• B. $75\times 2^9$
• C. $2^9$
• D. $45\times 2^9$

Answer 1

Answer: (D)

$\sum _{r=1}^{10}r(r-1{)}^{10}{C}_r$
$=\sum _{r=1}^{10}r(r-1)\frac{10!}{r(r-1)(r-2)!(10-r)!}$
$=90\sum _{r=2}^{10}{}^8{C}_{r-2}$
$=90\times 2^8=45\times 2^9$