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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

(Σ_(a=0) ^4 C_a ^4 x^(2a) )(Σ_(b=0) ^7 C_b ^7 x^(3b) )(Σ_(c=0) ^(12) C_c ^(12) x^(4c) )  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=0  C_0 ^4 C_1 ^7 C_2 ^(12) +C_1 ^4 C_3 ^7 C_0 ^(12) +C_2 ^4 C_1 ^7 C_1 ^(12) +C_4 ^4 C_1 ^7 C_0 ^(12)   =1113 ✓

$$\left(\underset{{a}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}{C}_{{a}} ^{\mathrm{4}} {x}^{\mathrm{2}{a}} \right)\left(\underset{{b}=\mathrm{0}} {\overset{\mathrm{7}} {\sum}}{C}_{{b}} ^{\mathrm{7}} {x}^{\mathrm{3}{b}} \right)\left(\underset{{c}=\mathrm{0}} {\overset{\mathrm{12}} {\sum}}{C}_{{c}} ^{\mathrm{12}} {x}^{\mathrm{4}{c}} \right) \\ $$$$\mathrm{2}{a}+\mathrm{3}{b}+\mathrm{4}{c}=\mathrm{11} \\ $$$$\Rightarrow{a}=\mathrm{0},\:{b}=\mathrm{1},\:{c}=\mathrm{2} \\ $$$$\Rightarrow{a}=\mathrm{1},\:{b}=\mathrm{3},\:{c}=\mathrm{0} \\ $$$$\Rightarrow{a}=\mathrm{2},\:{b}=\mathrm{1},\:{c}=\mathrm{1} \\ $$$$\Rightarrow{a}=\mathrm{4},\:{b}=\mathrm{1},\:{c}=\mathrm{0} \\ $$$${C}_{\mathrm{0}} ^{\mathrm{4}} {C}_{\mathrm{1}} ^{\mathrm{7}} {C}_{\mathrm{2}} ^{\mathrm{12}} +{C}_{\mathrm{1}} ^{\mathrm{4}} {C}_{\mathrm{3}} ^{\mathrm{7}} {C}_{\mathrm{0}} ^{\mathrm{12}} +{C}_{\mathrm{2}} ^{\mathrm{4}} {C}_{\mathrm{1}} ^{\mathrm{7}} {C}_{\mathrm{1}} ^{\mathrm{12}} +{C}_{\mathrm{4}} ^{\mathrm{4}} {C}_{\mathrm{1}} ^{\mathrm{7}} {C}_{\mathrm{0}} ^{\mathrm{12}} \\ $$$$=\mathrm{1113}\:\checkmark \\ $$

Commented by manxsol last updated on 31/Dec/22

today learn,thanks Sir W

$${today}\:{learn},{thanks}\:{Sir}\:{W} \\ $$

Commented by Michaelfaraday last updated on 31/Dec/22

thanks

$${thanks} \\ $$

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