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Question Number 160909 by naka3546 last updated on 09/Dec/21
How  many  3−digits  number  such  that  sum  of  its  digits  is  11 ?
$${How}\:\:{many}\:\:\mathrm{3}−{digits}\:\:{number}\:\:{such}\:\:{that}\:\:{sum}\:\:{of}\:\:{its}\:\:{digits}\:\:{is}\:\:\mathrm{11}\:? \\ $$
Answered by mr W last updated on 09/Dec/21
METHOD 1  (x+x^2 +x^3 +...x^9 )(1+x+x^2 +...+x^9 )^2   =((x(1−x^9 )(1−x^(10) )^2 )/((1−x)^3 ))  =x(1−x^9 −2x^(10) +2x^(19) +x^(20) −x^(29) )Σ_(k=0) ^∞ C_2 ^(k+2) x^k   coef. of x^(11)  is  C_2 ^(12) −C_2 ^3 −2×C_2 ^2 =66−3−2=61  that means there are 61 such 3−digit  numbers.
$$\boldsymbol{{METHOD}}\:\mathrm{1} \\ $$$$\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…{x}^{\mathrm{9}} \right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…+{x}^{\mathrm{9}} \right)^{\mathrm{2}} \\ $$$$=\frac{{x}\left(\mathrm{1}−{x}^{\mathrm{9}} \right)\left(\mathrm{1}−{x}^{\mathrm{10}} \right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} } \\ $$$$={x}\left(\mathrm{1}−{x}^{\mathrm{9}} −\mathrm{2}{x}^{\mathrm{10}} +\mathrm{2}{x}^{\mathrm{19}} +{x}^{\mathrm{20}} −{x}^{\mathrm{29}} \right)\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{2}} ^{{k}+\mathrm{2}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{11}} \:{is} \\ $$$${C}_{\mathrm{2}} ^{\mathrm{12}} −{C}_{\mathrm{2}} ^{\mathrm{3}} −\mathrm{2}×{C}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{66}−\mathrm{3}−\mathrm{2}=\mathrm{61} \\ $$$${that}\:{means}\:{there}\:{are}\:\mathrm{61}\:{such}\:\mathrm{3}−{digit} \\ $$$${numbers}. \\ $$
Commented by Tawa11 last updated on 09/Dec/21
Great sir.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$
Answered by mr W last updated on 09/Dec/21
METHOD 2  abc with a+b+c=11  a=1:  b+c=10 ⇒9 numbers  a=2:  b+c=9 ⇒10 numbers  a=3:  b+c=8 ⇒9 numbers  a=4:  b+c=7 ⇒8 numbers  ...  a=9:  b+c=2 ⇒3 numbers  total: 9+10+9+8+...+3=61
$$\boldsymbol{{METHOD}}\:\mathrm{2} \\ $$$${abc}\:{with}\:{a}+{b}+{c}=\mathrm{11} \\ $$$${a}=\mathrm{1}:\:\:{b}+{c}=\mathrm{10}\:\Rightarrow\mathrm{9}\:{numbers} \\ $$$${a}=\mathrm{2}:\:\:{b}+{c}=\mathrm{9}\:\Rightarrow\mathrm{10}\:{numbers} \\ $$$${a}=\mathrm{3}:\:\:{b}+{c}=\mathrm{8}\:\Rightarrow\mathrm{9}\:{numbers} \\ $$$${a}=\mathrm{4}:\:\:{b}+{c}=\mathrm{7}\:\Rightarrow\mathrm{8}\:{numbers} \\ $$$$… \\ $$$${a}=\mathrm{9}:\:\:{b}+{c}=\mathrm{2}\:\Rightarrow\mathrm{3}\:{numbers} \\ $$$${total}:\:\mathrm{9}+\mathrm{10}+\mathrm{9}+\mathrm{8}+…+\mathrm{3}=\mathrm{61} \\ $$
Commented by naka3546 last updated on 09/Dec/21
Thank  you,  sir.
$${Thank}\:\:{you},\:\:{sir}.\: \\ $$

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