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Question:

The value of (21C1+10C1) + (21C2+10C2) + (21C3+10C3) + (21C4+10C4) + ... + (21C10+10C10) is:

220ð•’¶10

221ð•’¶11

221ð•’¶10

220ð•’¶9

Solution:

(21C1𕒵0C1)+(21C2𕒵0C2)+(21C3𕒵0C3)+(21C4+10C4)+..+(21C10𕒵0C10)=21C1+21C2+21C3+.+21C10−[210𕒵]=12[221C1+221C2+221C3+.+221C10]−[210𕒵]=12[21C1+21C2++21C10+21C11+.+21C19+21C20]−[210𕒵]=12[221𕒵𕒵]−[210𕒵]=220𕒵𕒶10+1=220𕒶10