Question

Find the value of: ${{}^{20}C}_0-{{}^{20}C}_1+{{}^{20}C}_2-.....+{{}^{20}C}_{18}=$

• A. $1$
• B. $0$
• C. $19$
• D. $20$

Answer 1

The correct option is C $19$

${}^{20}{C}_0-{}^{20}{C}_1+{}^{20}{C}_2-\dots +{}^{20}{C}_{18}=$

Since$(1+x{)}^{20}={}^{20}{C}_0+{}^{20}{C}_{1}x+\dots {+}^{20}{C}_{20}{x}^{20}$

Put $x=-1\\$

${}^{20}{C}_0{-}^{20}{C}_1+\dots +{}^{20}{C}_{18}-{}^{20}{C}_{19}+{}^{20}{C}_{20}$

$=[1+(-1){]}^{20}=0$

$\therefore {}^{20}{C}_{20}-{}^{20}{C}_{19}=-19$

${}^{20}{C}_0-{}^{20}{C}_1+{}^{20}{C}_2+(-1{)}^{20}{C}_3+\dots +{}^{20}{C}_{18}=19$