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Question Number 25786 by gugalesweta@gmail.com last updated on 14/Dec/17

If C_0 + C_1 + C_2 +...+ C_n  = 256, then  ^(2n) C_2  is equal to

$$\mathrm{If}\:{C}_{\mathrm{0}} +\:{C}_{\mathrm{1}} +\:{C}_{\mathrm{2}} +...+\:{C}_{{n}} \:=\:\mathrm{256},\:\mathrm{then} \\ $$$$\:^{\mathrm{2}{n}} {C}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Answered by mrW1 last updated on 14/Dec/17

C_0 +C_1 +...+C_n =2^n =256=2^8   ⇒n=8  C_2 ^(2n) =C_2 ^(16) =((16×15)/(2×1))=120

$${C}_{\mathrm{0}} +{C}_{\mathrm{1}} +...+{C}_{{n}} =\mathrm{2}^{{n}} =\mathrm{256}=\mathrm{2}^{\mathrm{8}} \\ $$$$\Rightarrow{n}=\mathrm{8} \\ $$$${C}_{\mathrm{2}} ^{\mathrm{2}{n}} ={C}_{\mathrm{2}} ^{\mathrm{16}} =\frac{\mathrm{16}×\mathrm{15}}{\mathrm{2}×\mathrm{1}}=\mathrm{120} \\ $$

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