Question

${}^{20}{C}_4+2\cdot {}^{20}{C}_3+{}^{20}{C}_2-{}^{22}{C}_{18}$ is equal to :

• A. 0
• B. 1242
• C. 7315
• D. 6345

Answer 1

Answer: (A)

If $n$ and $r$ positive integer and $0\le r<n$ ,
then ${}^n{C}_r+{}^n{C}_{r-1}={}^{n+1}{C}_r$
$\therefore {}^{20}{C}_4+2.{}^{20}{C}_3+{}^{20}{C}_2-{}^{22}{C}_{18}$
$=({}^{20}{C}_4+{}^{30}{C}_3)+({}^{20}{C}_3+{}^{20}{C}_2)-{}^{22}{C}_{18}$
$={}^{21}{C}_4+{}^{21}{C}_3-{}^{22}{C}_{18}$
$={}^{22}{C}_4-{}^{22}{C}_{18}$
$={}^{22}{C}_{18}-{}^{22}{C}_{18}$
$=0$