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Question Number 183843 by Michaelfaraday last updated on 30/Dec/22
Answered by mr W last updated on 31/Dec/22
(∑4a=0Ca4x2a)(∑7b=0Cb7x3b)(∑12c=0Cc12x4c)2a+3b+4c=11⇒a=0,b=1,c=2⇒a=1,b=3,c=0⇒a=2,b=1,c=1⇒a=4,b=1,c=0C04C17C212+C14C37C012+C24C17C112+C44C17C012=1113✓
Commented by manxsol last updated on 31/Dec/22
todaylearn,thanksSirW
Commented by Michaelfaraday last updated on 31/Dec/22
thanks
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