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Question Number 57594 by naka3546 last updated on 08/Apr/19

ABCD  is  four  digits  integers .  How  many  ABCD  that  suitable  with  A+B+C+D  =  25 ?

$${ABCD}\:\:{is}\:\:{four}\:\:{digits}\:\:{integers}\:. \\ $$$${How}\:\:{many}\:\:{ABCD}\:\:{that}\:\:{suitable}\:\:{with}\:\:{A}+{B}+{C}+{D}\:\:=\:\:\mathrm{25}\:? \\ $$

Answered by mr W last updated on 08/Apr/19

A: 1,2,..,9  B,C,D: 0,1,2,..,9  A+B+C+D=25  (x+x^2 +x^3 +...+x^9 )(1+x+x^2 +x^3 +...+x^9 )^3   =((x(1−x^9 )(1−x^(10) )^3 )/((1−x)^4 ))  =((x(1−x^9 )(1−3x^(10) +3x^(20) −x^(30) ))/((1−x)^4 ))  =((x(1−x^9 −3x^(10) +3x^(19) +3x^(20) +...))/((1−x)^4 ))  =x(1−x^9 −3x^(10) +3x^(19) +3x^(20) +...)Σ_(k=0) ^∞ C_3 ^(k+3) x^k   coef. of x^(25)  is  C_3 ^(27) −C_3 ^(18) −3C_3 ^(17) +3C_3 ^8 +3C_3 ^7 =2925−816−3×680+3×56+3×35=342  ⇒there are 342 suitable numbers.

$${A}:\:\mathrm{1},\mathrm{2},..,\mathrm{9} \\ $$$${B},{C},{D}:\:\mathrm{0},\mathrm{1},\mathrm{2},..,\mathrm{9} \\ $$$${A}+{B}+{C}+{D}=\mathrm{25} \\ $$$$\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...+{x}^{\mathrm{9}} \right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...+{x}^{\mathrm{9}} \right)^{\mathrm{3}} \\ $$$$=\frac{{x}\left(\mathrm{1}−{x}^{\mathrm{9}} \right)\left(\mathrm{1}−{x}^{\mathrm{10}} \right)^{\mathrm{3}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} } \\ $$$$=\frac{{x}\left(\mathrm{1}−{x}^{\mathrm{9}} \right)\left(\mathrm{1}−\mathrm{3}{x}^{\mathrm{10}} +\mathrm{3}{x}^{\mathrm{20}} −{x}^{\mathrm{30}} \right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} } \\ $$$$=\frac{{x}\left(\mathrm{1}−{x}^{\mathrm{9}} −\mathrm{3}{x}^{\mathrm{10}} +\mathrm{3}{x}^{\mathrm{19}} +\mathrm{3}{x}^{\mathrm{20}} +...\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} } \\ $$$$={x}\left(\mathrm{1}−{x}^{\mathrm{9}} −\mathrm{3}{x}^{\mathrm{10}} +\mathrm{3}{x}^{\mathrm{19}} +\mathrm{3}{x}^{\mathrm{20}} +...\right)\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{3}} ^{{k}+\mathrm{3}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{25}} \:{is} \\ $$$${C}_{\mathrm{3}} ^{\mathrm{27}} −{C}_{\mathrm{3}} ^{\mathrm{18}} −\mathrm{3}{C}_{\mathrm{3}} ^{\mathrm{17}} +\mathrm{3}{C}_{\mathrm{3}} ^{\mathrm{8}} +\mathrm{3}{C}_{\mathrm{3}} ^{\mathrm{7}} =\mathrm{2925}−\mathrm{816}−\mathrm{3}×\mathrm{680}+\mathrm{3}×\mathrm{56}+\mathrm{3}×\mathrm{35}=\mathrm{342} \\ $$$$\Rightarrow{there}\:{are}\:\mathrm{342}\:{suitable}\:{numbers}. \\ $$

Commented by mr W last updated on 08/Apr/19

for more details see Q21800.

$${for}\:{more}\:{details}\:{see}\:{Q}\mathrm{21800}. \\ $$

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