Question

The value of ${}^{40}{C}_{31}+\sum _{j=0}^{10}{}^{40+j}{C}_{10+j}$ is equal to

• A. ${}^{51}{C}_{20}$
• B. $2{.}^{50}{C}_{20}$
• C. $2{.}^{45}{C}_{15}$
• D. none of these

Answer 1

The correct option is A ${}^{51}{C}_{20}$

Expanding, we get

We know that,

${}^n{C}_{r-1}{+}^n{C}_r{=}^{n+1}{C}_r$

Therefore,

${}^{40}{C}_{31}+[{}^{40}{C}_{10}+{}^{41}{C}_{11}+{}^{42}{C}_{12}+...{}^{50}{C}_{20}]$
$={}^{40}{C}_9+[{}^{40}{C}_{10}+{}^{41}{C}_{11}+{}^{42}{C}_{12}+...{}^{50}{C}_{20}]$
$=({}^{40}{C}_9+{}^{40}{C}_{10})+{}^{41}{C}_{11}+{}^{42}{C}_{12}+...{}^{50}{C}_{20}$
$=({}^{41}{C}_{10}+{}^{41}{C}_{11})+{}^{42}{C}_{12}+...{}^{50}{C}_{20}$
$={}^{42}{C}_{11}+{}^{42}{C}_{12}+...{}^{50}{C}_{20}$
:
:
:
$={}^{49}{C}_{18}+{}^{49}{C}_{19}+{}^{50}{C}_{20}$
$={}^{50}{C}_{19}+{}^{50}{C}_{20}$
$={}^{51}{C}_{20}$