The correct option is D 220−210
(21C1− 10C1)+(21C2− 10C2)+(21C3− 10C3)+⋯+(21C10− 10C10)
=(21C1+ 21C2+⋯+ 21C10)−(10C1+ 10C2+⋯+ 10C10)
Now, 21C1+ 21C2+⋯+ 21C10
=12(21C1+ 21C2+⋯+ 21C10+ 21C1+21C2+⋯+ 21C10)
=12(21C1+ 21C2+⋯+ 21C10+ 21C20+21C19+⋯+ 21C11)
=12(221−2)=220−1
10C1+ 10C2+⋯+ 10C10
=210−1
∴ Required sum =220−210