Question and Answers Forum

All Questions      Topic List

Permutation and Combination Questions

Previous in All Question      Next in All Question      

Previous in Permutation and Combination      Next in Permutation and Combination      

Question Number 102816 by mr W last updated on 11/Jul/20

How many 6 digit numbers exist whose digits have exactly the sum 13? for example 120505 is such a number.

$${How}\:{many}\:\mathrm{6}\:{digit}\:{numbers}\:{exist} \\ $$$${whose}\:{digits}\:{have}\:{exactly}\:{the}\:{sum}\:\mathrm{13}? \\ $$$$ \\ $$$${for}\:{example}\:\mathrm{120505}\:{is}\:{such}\:{a}\:{number}. \\ $$

Commented by Rasheed.Sindhi last updated on 11/Jul/20

Universal Set {0,0,0,0,1,1,1,1,1,2,2,2,2,2,3,3,3,3, 4,4,4,5,5,6,6,7,8,9}

$$\mathrm{Universal}\:\mathrm{Set} \\ $$$$\left\{\mathrm{0},\mathrm{0},\mathrm{0},\mathrm{0},\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{2},\mathrm{2},\mathrm{2},\mathrm{2},\mathrm{2},\mathrm{3},\mathrm{3},\mathrm{3},\mathrm{3},\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4},\mathrm{4},\mathrm{4},\mathrm{5},\mathrm{5},\mathrm{6},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right\} \\ $$

Commented by aurpeyz last updated on 11/Jul/20

660100 508000 309000...it will be much o. how should this be[done?

$$\mathrm{660100}\:\mathrm{508000}\:\mathrm{309000}...\mathrm{it}\:\mathrm{will}\:\mathrm{be}\:\mathrm{much}\:\mathrm{o}.\:\mathrm{how}\:\mathrm{should}\:\mathrm{this}\:\mathrm{be}\left[\mathrm{done}?\right. \\ $$

Commented by mr W last updated on 11/Jul/20

Rasheed sir: i don′t know much about set methods. can you give some explanation and how to arrive at the result? thanks!

$${Rasheed}\:{sir}:\:{i}\:{don}'{t}\:{know}\:{much} \\ $$$${about}\:{set}\:{methods}.\:{can}\:{you}\:{give}\:{some} \\ $$$${explanation}\:{and}\:{how}\:{to}\:{arrive}\:{at}\:{the} \\ $$$${result}?\:{thanks}! \\ $$

Commented by mr W last updated on 11/Jul/20

i see, thanks! do you have other ideas?

$${i}\:{see},\:{thanks}!\:{do}\:{you}\:{have}\:{other}\:{ideas}? \\ $$

Commented by Rasheed.Sindhi last updated on 11/Jul/20

Sir, actually it′s not much helpfull,I think now.I only extended the set{0,1,2,...,9} by writing maximum possible occurigs of each digit in the required numbers.

$${Sir},\:{actually}\:{it}'{s}\:{not}\:{much} \\ $$$${helpfull},{I}\:{think}\:{now}.{I}\:{only}\:\: \\ $$$${extended}\:{the}\:{set}\left\{\mathrm{0},\mathrm{1},\mathrm{2},...,\mathrm{9}\right\}\:{by} \\ $$$${writing}\:{maximum}\:{possible}\: \\ $$$${occurigs}\:{of}\:{each}\:{digit}\:{in}\:{the}\: \\ $$$${required}\:{numbers}. \\ $$

Commented by mr W last updated on 11/Jul/20

i got 6027 such numbers.

$${i}\:{got}\:\mathrm{6027}\:{such}\:{numbers}. \\ $$

Commented by Rasheed.Sindhi last updated on 11/Jul/20

Sir I thought to attack the problem as follows: (i) To determine the number of partions(in an order) of 13 using above ′universal set′ {0,0,0,0,1,1,1,1,1,2,2,2,2,2,3,3,3,3, 4,4,4,5,5,6,6,7,8,9} (ii)To determine number of ′permutations′ of each partitions. (iii)After than number of excluding unwanted numbers. I am not confident of these ideas.You can do better.

$${Sir}\:{I}\:{thought}\:{to}\:{attack}\:{the} \\ $$$${problem}\:{as}\:{follows}: \\ $$$$\left({i}\right)\:\mathcal{T}{o}\:{determine}\:{the}\:{number} \\ $$$${of}\:{partions}\left({in}\:{an}\:{order}\right) \\ $$$${of}\:\mathrm{13}\:{using}\:{above}\:'{universal}\:{set}' \\ $$$$\left\{\mathrm{0},\mathrm{0},\mathrm{0},\mathrm{0},\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{2},\mathrm{2},\mathrm{2},\mathrm{2},\mathrm{2},\mathrm{3},\mathrm{3},\mathrm{3},\mathrm{3},\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\mathrm{4},\mathrm{4},\mathrm{4},\mathrm{5},\mathrm{5},\mathrm{6},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right\} \\ $$$$\left({ii}\right)\mathcal{T}{o}\:{determine}\:{number}\:{of} \\ $$$$'{permutations}'\:{of}\:{each} \\ $$$${partitions}. \\ $$$$\left({iii}\right){After}\:{than}\:{number}\:{of} \\ $$$${excluding}\:{unwanted}\:{numbers}. \\ $$$$\:\:\:\:\:{I}\:{am}\:{not}\:{confident}\:{of}\:{these} \\ $$$${ideas}.{You}\:{can}\:{do}\:{better}. \\ $$

Commented by prakash jain last updated on 11/Jul/20

This method i think is known as generating function method, used in problem involving coins etc.

Commented by prakash jain last updated on 11/Jul/20

Thanks. I will recheck calculation for both methods.

$$\mathrm{Thanks}.\:\mathrm{I}\:\mathrm{will}\:\mathrm{recheck}\:\mathrm{calculation} \\ $$$$\mathrm{for}\:\mathrm{both}\:\mathrm{methods}. \\ $$

Commented by PRITHWISH SEN 2 last updated on 11/Jul/20

Sir itis really hard . But I might find a different way. For instance let find the 3 digits numbers in between 100 and 200 whose digit sum is 12 Sum of digits Numbers 1 −−−−100 2−−−−101−−−−110 3−−−−102−−−−111−−−−120 4−−−−103−−−−112−−−−121−−−−130 5−−−−104−−−−113−−−−122−−−−131 6−−−−105−−−−114−−−−123−−−−132 7−−−−106−−−−115−−−−124−−−−133 8−−−−107−−−−116−−−−125−−−−134 9−−−−108−−−−117−−−−126−−−−135 10−−− 109−−−−118−−−− 127−−−−136 11−−−−−−−−−119−−−−128−−−−137 12−−−−−−−−−−−−−−−129−−−−138 and so on.. so the first number is 129 and the last number is 129+9×n<200 n=7 ∴ no. of 3 digits number whose sum is 12 is = 7+1=8 sir is it can be helpful for such problems ? Sir Rasheed and Mr.Wsir please share your comments.

$$\mathrm{Sir}\:\mathrm{itis}\:\mathrm{really}\:\mathrm{hard}\:.\:\mathrm{But}\:\mathrm{I}\:\mathrm{might}\:\mathrm{find}\:\mathrm{a}\:\mathrm{different} \\ $$$$\mathrm{way}. \\ $$$$\mathrm{For}\:\mathrm{instance}\:\mathrm{let}\:\mathrm{find}\:\mathrm{the}\:\mathrm{3}\:\mathrm{digits}\:\mathrm{numbers}\:\mathrm{in}\:\mathrm{between} \\ $$$$\mathrm{100}\:\mathrm{and}\:\mathrm{200}\:\mathrm{whose}\:\mathrm{digit}\:\mathrm{sum}\:\mathrm{is}\:\mathrm{12}\: \\ $$$$\boldsymbol{\mathrm{Sum}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{digits}}\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Numbers}} \\ $$$$\mathrm{1}\:−−−−\mathrm{100} \\ $$$$\mathrm{2}−−−−\mathrm{101}−−−−\mathrm{110} \\ $$$$\mathrm{3}−−−−\mathrm{102}−−−−\mathrm{111}−−−−\mathrm{120} \\ $$$$\mathrm{4}−−−−\mathrm{103}−−−−\mathrm{112}−−−−\mathrm{121}−−−−\mathrm{130} \\ $$$$\mathrm{5}−−−−\mathrm{104}−−−−\mathrm{113}−−−−\mathrm{122}−−−−\mathrm{131} \\ $$$$\mathrm{6}−−−−\mathrm{105}−−−−\mathrm{114}−−−−\mathrm{123}−−−−\mathrm{132} \\ $$$$\mathrm{7}−−−−\mathrm{106}−−−−\mathrm{115}−−−−\mathrm{124}−−−−\mathrm{133} \\ $$$$\mathrm{8}−−−−\mathrm{107}−−−−\mathrm{116}−−−−\mathrm{125}−−−−\mathrm{134} \\ $$$$\mathrm{9}−−−−\mathrm{108}−−−−\mathrm{117}−−−−\mathrm{126}−−−−\mathrm{135} \\ $$$$\mathrm{10}−−−\:\mathrm{109}−−−−\mathrm{118}−−−−\:\mathrm{127}−−−−\mathrm{136} \\ $$$$\mathrm{11}−−−−−−−−−\mathrm{119}−−−−\mathrm{128}−−−−\mathrm{137} \\ $$$$\mathrm{12}−−−−−−−−−−−−−−−\mathrm{129}−−−−\mathrm{138} \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{so}}\:\boldsymbol{\mathrm{on}}.. \\ $$$$\boldsymbol{\mathrm{so}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{first}}\:\boldsymbol{\mathrm{number}}\:\boldsymbol{\mathrm{is}}\:\mathrm{129}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{last}}\:\boldsymbol{\mathrm{number}}\: \\ $$$$\boldsymbol{\mathrm{is}}\:\mathrm{129}+\mathrm{9}×\boldsymbol{\mathrm{n}}<\mathrm{200} \\ $$$$\boldsymbol{\mathrm{n}}=\mathrm{7} \\ $$$$\therefore\:\mathrm{no}.\:\mathrm{of}\:\mathrm{3}\:\mathrm{digits}\:\mathrm{number}\:\mathrm{whose}\:\mathrm{sum}\:\mathrm{is}\:\mathrm{12}\:\mathrm{is} \\ $$$$=\:\mathrm{7}+\mathrm{1}=\mathrm{8} \\ $$$$\mathrm{sir}\:\mathrm{is}\:\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{helpful}\:\mathrm{for}\:\mathrm{such}\:\mathrm{problems}\:? \\ $$$$\mathrm{Sir}\:\mathrm{Rasheed}\:\mathrm{and}\:\mathrm{Mr}.\mathrm{Wsir}\:\mathrm{please}\:\mathrm{share}\:\mathrm{your} \\ $$$$\mathrm{comments}. \\ $$

Commented by mr W last updated on 11/Jul/20

yes! this is the best method to solve. please recheck your formula, i think it contains an error. i got: C_5 ^(17) −C_5 ^8 −5C_5 ^7 =6027

$${yes}!\:{this}\:{is}\:{the}\:{best}\:{method}\:{to}\:{solve}. \\ $$$${please}\:{recheck}\:{your}\:{formula},\:{i}\:{think} \\ $$$${it}\:{contains}\:{an}\:{error}.\:{i}\:{got}: \\ $$$${C}_{\mathrm{5}} ^{\mathrm{17}} −{C}_{\mathrm{5}} ^{\mathrm{8}} −\mathrm{5}{C}_{\mathrm{5}} ^{\mathrm{7}} =\mathrm{6027} \\ $$

Commented by mr W last updated on 11/Jul/20

big thanks to all for the many ideas!

$${big}\:{thanks}\:{to}\:{all}\:{for}\:{the}\:{many}\:{ideas}! \\ $$

Commented by prakash jain last updated on 11/Jul/20

See Q22040

$$\mathrm{See}\:\mathrm{Q22040} \\ $$

Commented by prakash jain last updated on 11/Jul/20

Q55502

$$\mathrm{Q55502} \\ $$

Commented by prakash jain last updated on 11/Jul/20

Several other questions answered by mr W only using method. mr W, is the same method not applicable here. I did remember some previous questions so searched.

$$\mathrm{Several}\:\mathrm{other}\:\mathrm{questions}\:\mathrm{answered} \\ $$$$\mathrm{by}\:\mathrm{mr}\:\mathrm{W}\:\mathrm{only}\:\mathrm{using}\:\mathrm{method}. \\ $$$$\mathrm{mr}\:\mathrm{W},\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{method}\:\mathrm{not} \\ $$$$\mathrm{applicable}\:\mathrm{here}. \\ $$$$\mathrm{I}\:\mathrm{did}\:\mathrm{remember}\:\mathrm{some}\:\mathrm{previous}\: \\ $$$$\mathrm{questions}\:\mathrm{so}\:\mathrm{searched}. \\ $$

Commented by PRITHWISH SEN 2 last updated on 11/Jul/20

Thank you sirs

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sirs} \\ $$

Commented by mr W last updated on 11/Jul/20

there is no standard way for such problems. the way you are trying can reach the goal, but it could be tough.

$${there}\:{is}\:{no}\:{standard}\:{way}\:{for}\:{such} \\ $$$${problems}.\:{the}\:{way}\:{you}\:{are}\:{trying}\:{can} \\ $$$${reach}\:{the}\:{goal},\:{but}\:{it}\:{could}\:{be}\:{tough}. \\ $$

Commented by Rasheed.Sindhi last updated on 11/Jul/20

Sir an idea comes to me! If sum of digits is 13 that means numbers which leaves remainder 1 when they′re divided by 3. That is 3k+1 type numbers. The numbers leaves remainder 4 when they′re divided by 9. That is the numbers of 9k+4 types....Some extra numbers ....

$${Sir}\:{an}\:{idea}\:{comes}\:{to}\:{me}!\:{If}\:{sum} \\ $$$${of}\:{digits}\:{is}\:\mathrm{13}\:{that}\:{means} \\ $$$${numbers}\:{which}\:{leaves} \\ $$$${remainder}\:\mathrm{1}\:{when}\:{they}'{re} \\ $$$${divided}\:{by}\:\mathrm{3}.\:{That}\:{is}\:\mathrm{3}{k}+\mathrm{1}\:{type} \\ $$$${numbers}. \\ $$$$\:\mathcal{T}{he}\:{numbers}\:{leaves}\:{remainder} \\ $$$$\mathrm{4}\:{when}\:{they}'{re}\:{divided}\:{by}\:\mathrm{9}. \\ $$$$\mathcal{T}{hat}\:{is}\:{the}\:{numbers}\:{of}\:\mathrm{9}{k}+\mathrm{4} \\ $$$${types}....{Some}\:{extra}\:{numbers} \\ $$$$.... \\ $$

Commented by PRITHWISH SEN 2 last updated on 11/Jul/20

sir but the number 300001 =3×100000+1 but 3+0+0+0+0+1≠13 and 3×94528+1=283585≠13

$$\mathrm{sir}\:\mathrm{but}\:\mathrm{the}\:\mathrm{number} \\ $$$$\mathrm{300001}\:=\mathrm{3}×\mathrm{100000}+\mathrm{1} \\ $$$$\mathrm{but}\:\mathrm{3}+\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{1}\neq\mathrm{13} \\ $$$$\mathrm{and} \\ $$$$\mathrm{3}×\mathrm{94528}+\mathrm{1}=\mathrm{283585}\neq\mathrm{13} \\ $$

Commented by Rasheed.Sindhi last updated on 11/Jul/20

Hello sir PRITHWISH SEN! You′re right sir! Some extra numbers will also included.So...

$${Hello}\:{sir}\:{PRITHWISH}\:{SEN}! \\ $$$${You}'{re}\:{right}\:{sir}!\:{Some}\:{extra} \\ $$$${numbers}\:{will}\:{also}\:{included}.{So}... \\ $$

Commented by PRITHWISH SEN 2 last updated on 11/Jul/20

Please check this 1^(st) Digit No. of numbers 9 The coefficient of 4 in (a^0 +a^1 +a^2 +a^3 +a^4 +a^5 +a^6 +a^7 +a^8 +a^9 )^5 8 The coeffi. of 5 in (a^0 +a^1 +a^2 +a^3 +a^4 +a^5 +a^6 +a^7 +a^8 +a^9 )^5 7 The coeffi. of 6 in (a^0 +a^1 +a^2 +a^3 +a^4 +a^5 +a^6 +a^7 +a^8 +a^9 )^5 6 The coeffi. of 7 in (a^0 +a^1 +a^2 +a^3 +a^4 +a^5 +a^6 +a^7 +a^8 +a^9 )^5 5 The coeffi. of 8 in (a^0 +a^1 +a^2 +a^3 +a^4 +a^5 +a^6 +a^7 +a^8 +a^9 )^5 4 The coeffi. of 9 in (a^0 +a^1 +a^2 +a^3 +a^4 +a^5 +a^6 +a^7 +a^8 +a^9 )^5 3 The coeffi. of 10 in (a^0 +a^1 +a^2 +a^3 +a^4 +a^5 +a^6 +a^7 +a^8 +a^9 )^5 2 The coeffi. of 11 in (a^0 +a^1 +a^2 +a^3 +a^4 +a^5 +a^6 +a^7 +a^8 +a^9 )^5 1 The coeffi. of 12 in (a^0 +a^1 +a^2 +a^3 +a^4 +a^5 +a^6 +a^7 +a^8 +a^9 )^5

$$\mathrm{Please}\:\mathrm{check}\:\mathrm{this} \\ $$$$\:\mathrm{1}^{\boldsymbol{\mathrm{st}}} \:\boldsymbol{\mathrm{Digit}}\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{No}}.\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{numbers}} \\ $$$$\:\:\:\:\mathrm{9}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{coefficient}}\:\boldsymbol{\mathrm{of}}\:\mathrm{4}\:\boldsymbol{\mathrm{in}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{a}}^{\mathrm{0}} +\boldsymbol{\mathrm{a}}^{\mathrm{1}} +\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{a}}^{\mathrm{4}} +\boldsymbol{\mathrm{a}}^{\mathrm{5}} +\boldsymbol{\mathrm{a}}^{\mathrm{6}} +\boldsymbol{\mathrm{a}}^{\mathrm{7}} +\boldsymbol{\mathrm{a}}^{\mathrm{8}} +\boldsymbol{\mathrm{a}}^{\mathrm{9}} \right)^{\mathrm{5}} \\ $$$$\:\:\:\mathrm{8}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{coeffi}}.\:\boldsymbol{\mathrm{of}}\:\mathrm{5}\:\boldsymbol{\mathrm{in}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{a}}^{\mathrm{0}} +\boldsymbol{\mathrm{a}}^{\mathrm{1}} +\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{a}}^{\mathrm{4}} +\boldsymbol{\mathrm{a}}^{\mathrm{5}} +\boldsymbol{\mathrm{a}}^{\mathrm{6}} +\boldsymbol{\mathrm{a}}^{\mathrm{7}} +\boldsymbol{\mathrm{a}}^{\mathrm{8}} +\boldsymbol{\mathrm{a}}^{\mathrm{9}} \right)^{\mathrm{5}} \\ $$$$\:\:\:\mathrm{7}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{coeffi}}.\:\boldsymbol{\mathrm{of}}\:\:\mathrm{6}\:\boldsymbol{\mathrm{in}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{a}}^{\mathrm{0}} +\boldsymbol{\mathrm{a}}^{\mathrm{1}} +\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{a}}^{\mathrm{4}} +\boldsymbol{\mathrm{a}}^{\mathrm{5}} +\boldsymbol{\mathrm{a}}^{\mathrm{6}} +\boldsymbol{\mathrm{a}}^{\mathrm{7}} +\boldsymbol{\mathrm{a}}^{\mathrm{8}} +\boldsymbol{\mathrm{a}}^{\mathrm{9}} \right)^{\mathrm{5}} \\ $$$$\:\:\:\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{coeffi}}.\:\boldsymbol{\mathrm{of}}\:\:\mathrm{7}\:\boldsymbol{\mathrm{in}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{a}}^{\mathrm{0}} +\boldsymbol{\mathrm{a}}^{\mathrm{1}} +\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{a}}^{\mathrm{4}} +\boldsymbol{\mathrm{a}}^{\mathrm{5}} +\boldsymbol{\mathrm{a}}^{\mathrm{6}} +\boldsymbol{\mathrm{a}}^{\mathrm{7}} +\boldsymbol{\mathrm{a}}^{\mathrm{8}} +\boldsymbol{\mathrm{a}}^{\mathrm{9}} \right)^{\mathrm{5}} \\ $$$$\:\:\:\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{coeffi}}.\:\boldsymbol{\mathrm{of}}\:\:\mathrm{8}\:\boldsymbol{\mathrm{in}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{a}}^{\mathrm{0}} +\boldsymbol{\mathrm{a}}^{\mathrm{1}} +\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{a}}^{\mathrm{4}} +\boldsymbol{\mathrm{a}}^{\mathrm{5}} +\boldsymbol{\mathrm{a}}^{\mathrm{6}} +\boldsymbol{\mathrm{a}}^{\mathrm{7}} +\boldsymbol{\mathrm{a}}^{\mathrm{8}} +\boldsymbol{\mathrm{a}}^{\mathrm{9}} \right)^{\mathrm{5}} \\ $$$$\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{coeffi}}.\:\boldsymbol{\mathrm{of}}\:\:\mathrm{9}\:\boldsymbol{\mathrm{in}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{a}}^{\mathrm{0}} +\boldsymbol{\mathrm{a}}^{\mathrm{1}} +\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{a}}^{\mathrm{4}} +\boldsymbol{\mathrm{a}}^{\mathrm{5}} +\boldsymbol{\mathrm{a}}^{\mathrm{6}} +\boldsymbol{\mathrm{a}}^{\mathrm{7}} +\boldsymbol{\mathrm{a}}^{\mathrm{8}} +\boldsymbol{\mathrm{a}}^{\mathrm{9}} \right)^{\mathrm{5}} \\ $$$$\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{coeffi}}.\:\boldsymbol{\mathrm{of}}\:\:\mathrm{10}\:\boldsymbol{\mathrm{in}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{a}}^{\mathrm{0}} +\boldsymbol{\mathrm{a}}^{\mathrm{1}} +\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{a}}^{\mathrm{4}} +\boldsymbol{\mathrm{a}}^{\mathrm{5}} +\boldsymbol{\mathrm{a}}^{\mathrm{6}} +\boldsymbol{\mathrm{a}}^{\mathrm{7}} +\boldsymbol{\mathrm{a}}^{\mathrm{8}} +\boldsymbol{\mathrm{a}}^{\mathrm{9}} \right)^{\mathrm{5}} \\ $$$$\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{coeffi}}.\:\boldsymbol{\mathrm{of}}\:\:\mathrm{11}\:\boldsymbol{\mathrm{in}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{a}}^{\mathrm{0}} +\boldsymbol{\mathrm{a}}^{\mathrm{1}} +\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{a}}^{\mathrm{4}} +\boldsymbol{\mathrm{a}}^{\mathrm{5}} +\boldsymbol{\mathrm{a}}^{\mathrm{6}} +\boldsymbol{\mathrm{a}}^{\mathrm{7}} +\boldsymbol{\mathrm{a}}^{\mathrm{8}} +\boldsymbol{\mathrm{a}}^{\mathrm{9}} \right)^{\mathrm{5}} \\ $$$$\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{coeffi}}.\:\boldsymbol{\mathrm{of}}\:\:\mathrm{12}\:\boldsymbol{\mathrm{in}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{a}}^{\mathrm{0}} +\boldsymbol{\mathrm{a}}^{\mathrm{1}} +\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{a}}^{\mathrm{4}} +\boldsymbol{\mathrm{a}}^{\mathrm{5}} +\boldsymbol{\mathrm{a}}^{\mathrm{6}} +\boldsymbol{\mathrm{a}}^{\mathrm{7}} +\boldsymbol{\mathrm{a}}^{\mathrm{8}} +\boldsymbol{\mathrm{a}}^{\mathrm{9}} \right)^{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Commented by prakash jain last updated on 11/Jul/20

First digits 1 to 9 rest all 0 to 9 (x^1 +....+x^9 )(1+...+x^9 )^5 Digits with sum 13 is given by coefficient of x^(13) in above expression. ((x(1−x^9 ))/(1−x))×(((1−x^(10) )^5 )/((1−x)^5 )) =x(1−x^9 −x^(10) +higherpower)(1−x)^(−6) =x(1−x^9 −5x^(10) )(1+Σ_(i=1) ^∞ ^(6+i−1) C_i x^i ) =(x−x^(10) −x^(11) )(...) Terms in ()that will given x^(13 ) x^(12) ,x^3 ,x^2 ^(17) C_5 −^8 C_3 −5^7 C_2 (correction (1−x^(10) )^5 was earlier written (1−x^(10) +..) it should be (1−5x^(10) +..)

$$\mathrm{First}\:\mathrm{digits}\:\mathrm{1}\:\mathrm{to}\:\mathrm{9} \\ $$$$\mathrm{rest}\:\mathrm{all}\:\mathrm{0}\:\mathrm{to}\:\mathrm{9} \\ $$$$\left({x}^{\mathrm{1}} +....+{x}^{\mathrm{9}} \right)\left(\mathrm{1}+...+{x}^{\mathrm{9}} \right)^{\mathrm{5}} \\ $$$$\mathrm{Digits}\:\mathrm{with}\:\mathrm{sum}\:\mathrm{13}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$$\mathrm{coefficient}\:\mathrm{of}\:{x}^{\mathrm{13}} \:\mathrm{in}\:\mathrm{above}\:\mathrm{expression}. \\ $$$$\frac{{x}\left(\mathrm{1}−{x}^{\mathrm{9}} \right)}{\mathrm{1}−{x}}×\frac{\left(\mathrm{1}−{x}^{\mathrm{10}} \right)^{\mathrm{5}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{5}} } \\ $$$$={x}\left(\mathrm{1}−{x}^{\mathrm{9}} −{x}^{\mathrm{10}} +\mathrm{higherpower}\right)\left(\mathrm{1}−{x}\right)^{−\mathrm{6}} \\ $$$$={x}\left(\mathrm{1}−{x}^{\mathrm{9}} −\mathrm{5}{x}^{\mathrm{10}} \right)\left(\mathrm{1}+\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\:^{\mathrm{6}+{i}−\mathrm{1}} {C}_{{i}} {x}^{{i}} \right) \\ $$$$=\left({x}−{x}^{\mathrm{10}} −{x}^{\mathrm{11}} \right)\left(...\right) \\ $$$$\mathrm{Terms}\:\mathrm{in}\:\left(\right)\mathrm{that}\:\mathrm{will}\:\mathrm{given}\:{x}^{\mathrm{13}\:} \: \\ $$$${x}^{\mathrm{12}} ,{x}^{\mathrm{3}} ,{x}^{\mathrm{2}} \\ $$$$\:\:^{\mathrm{17}} {C}_{\mathrm{5}} −^{\mathrm{8}} {C}_{\mathrm{3}} −\mathrm{5}\:^{\mathrm{7}} {C}_{\mathrm{2}} \\ $$$$\left(\mathrm{correction}\:\left(\mathrm{1}−{x}^{\mathrm{10}} \right)^{\mathrm{5}} \:\mathrm{was}\right. \\ $$$$\mathrm{earlier}\:\mathrm{written}\:\left(\mathrm{1}−{x}^{\mathrm{10}} +..\right) \\ $$$$\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\left(\mathrm{1}−\mathrm{5}{x}^{\mathrm{10}} +..\right) \\ $$

Commented by mr W last updated on 11/Jul/20

yes sir! i solved similar questions sometimes ago, but i can′t remember the old posts. how did you find these old posts?

$${yes}\:{sir}!\:{i}\:{solved}\:{similar}\:{questions} \\ $$$${sometimes}\:{ago},\:{but}\:{i}\:{can}'{t}\:{remember} \\ $$$${the}\:{old}\:{posts}.\:{how}\:{did}\:{you}\:{find}\:{these} \\ $$$${old}\:{posts}? \\ $$

Commented by prakash jain last updated on 11/Jul/20

I searched for generating or just generat etc. tried 2−3 search text.

$$\mathrm{I}\:\mathrm{searched}\:\mathrm{for}\:\mathrm{generating}\:\mathrm{or}\:\mathrm{just} \\ $$$$\mathrm{generat}\:\mathrm{etc}.\:\mathrm{tried}\:\mathrm{2}−\mathrm{3}\:\mathrm{search}\:\mathrm{text}. \\ $$

Commented by mr W last updated on 11/Jul/20

thanks sir! in this way one can find some old posts, great!

$${thanks}\:{sir}!\:{in}\:{this}\:{way}\:{one}\:{can}\:{find} \\ $$$${some}\:{old}\:{posts},\:{great}! \\ $$

Commented by PRITHWISH SEN 2 last updated on 11/Jul/20

sir prakash jain i think your 2^(nd) expression will be (1+....+x^9 )^5 not 6

$$\mathrm{sir}\:\mathrm{prakash}\:\mathrm{jain} \\ $$$$\mathrm{i}\:\mathrm{think}\:\mathrm{your}\:\mathrm{2}^{\mathrm{nd}} \mathrm{expression}\:\mathrm{will}\:\mathrm{be} \\ $$$$\left(\mathrm{1}+....+\mathrm{x}^{\mathrm{9}} \right)^{\mathrm{5}} \:\mathrm{not}\:\mathrm{6} \\ $$

Commented by prakash jain last updated on 11/Jul/20

yes. it is 5.

$${yes}.\:{it}\:{is}\:\mathrm{5}. \\ $$

Commented by prakash jain last updated on 11/Jul/20

ok. I realize now. in previous step i have written six.

$${ok}.\:{I}\:{realize}\:{now}.\:{in}\:{previous}\:{step} \\ $$$${i}\:{have}\:{written}\:{six}. \\ $$

Answered by mr W last updated on 11/Jul/20

let′s look at the general case: how many n digit numbers exist whose digits have exactly the sum m? say such a number is d_1 d_2 d_3 ...d_n with the conditions: 1≤d_1 ≤9 0≤d_2 ,d_3 ,...,d_n ≤9 so the question is how many integral solutions the following equation d_1 +d_2 +d_3 +...+d_n =m ...(I) has under the given conditions. we can use the generating functions for d_1 : x+x^2 +x^3 +...+x^9 for d_2 ,d_3 ,...,d_n : 1+x+x^2 +x^3 +...+x^9 for d_1 +d_2 +d_3 +...+d_n it is then (x+x^2 +x^3 +...+x^9 )(1+x+x^2 +x^3 +...+x^9 )^(n−1) =((x(1−x^9 )(1−x^(10) )^(n−1) )/((1−x)^n )) =x(1−x^9 )(1−x^(10) )^(n−1) Σ_(k=0) ^∞ C_(n−1) ^(k+n−1) x^k =x(1−x^9 )[Σ_(r=0) ^(n−1) (−1)^r C_r ^(n−1) x^(10r) ][Σ_(k=0) ^∞ C_(n−1) ^(k+n−1) x^k ] the coefficient of the x^m term in this generating function represents the number of solutions of eqn. (I). example: n=6, m=13 GF=x(1−x^9 )(1−5x^(10) +...)Σ_(k=0) ^∞ C_5 ^(k+5) x^k =x(1−x^9 −5x^(10) +...)Σ_(k=0) ^∞ C_5 ^(k+5) x^k coefficient of x^(13) term is (for k=12, 3, 2) C_5 ^(17) −C_5 ^8 −5C_5 ^7 =6027

$${let}'{s}\:{look}\:{at}\:{the}\:{general}\:{case}: \\ $$$${how}\:{many}\:{n}\:{digit}\:{numbers}\:{exist} \\ $$$${whose}\:{digits}\:{have}\:{exactly}\:{the}\:{sum}\:{m}? \\ $$$${say}\:{such}\:{a}\:{number}\:{is} \\ $$$${d}_{\mathrm{1}} {d}_{\mathrm{2}} {d}_{\mathrm{3}} ...{d}_{{n}} \\ $$$${with}\:{the}\:{conditions}: \\ $$$$\mathrm{1}\leqslant{d}_{\mathrm{1}} \leqslant\mathrm{9} \\ $$$$\mathrm{0}\leqslant{d}_{\mathrm{2}} ,{d}_{\mathrm{3}} ,...,{d}_{{n}} \leqslant\mathrm{9} \\ $$$${so}\:{the}\:{question}\:{is}\:{how}\:{many}\:{integral} \\ $$$${solutions}\:{the}\:{following}\:{equation} \\ $$$${d}_{\mathrm{1}} +{d}_{\mathrm{2}} +{d}_{\mathrm{3}} +...+{d}_{{n}} ={m}\:\:\:...\left({I}\right) \\ $$$${has}\:{under}\:{the}\:{given}\:{conditions}. \\ $$$${we}\:{can}\:{use}\:{the}\:{generating}\:{functions} \\ $$$${for}\:{d}_{\mathrm{1}} :\:{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...+{x}^{\mathrm{9}} \\ $$$${for}\:{d}_{\mathrm{2}} ,{d}_{\mathrm{3}} ,...,{d}_{{n}} :\:\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...+{x}^{\mathrm{9}} \\ $$$${for}\:{d}_{\mathrm{1}} +{d}_{\mathrm{2}} +{d}_{\mathrm{3}} +...+{d}_{{n}} \:{it}\:{is}\:{then} \\ $$$$\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...+{x}^{\mathrm{9}} \right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...+{x}^{\mathrm{9}} \right)^{{n}−\mathrm{1}} \\ $$$$=\frac{{x}\left(\mathrm{1}−{x}^{\mathrm{9}} \right)\left(\mathrm{1}−{x}^{\mathrm{10}} \right)^{{n}−\mathrm{1}} }{\left(\mathrm{1}−{x}\right)^{{n}} } \\ $$$$={x}\left(\mathrm{1}−{x}^{\mathrm{9}} \right)\left(\mathrm{1}−{x}^{\mathrm{10}} \right)^{{n}−\mathrm{1}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{{n}−\mathrm{1}} ^{{k}+{n}−\mathrm{1}} {x}^{{k}} \\ $$$$={x}\left(\mathrm{1}−{x}^{\mathrm{9}} \right)\left[\underset{{r}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{r}} {C}_{{r}} ^{{n}−\mathrm{1}} {x}^{\mathrm{10}{r}} \right]\left[\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{{n}−\mathrm{1}} ^{{k}+{n}−\mathrm{1}} {x}^{{k}} \right] \\ $$$${the}\:{coefficient}\:{of}\:{the}\:{x}^{{m}} \:{term}\:{in}\:{this} \\ $$$${generating}\:{function}\:{represents}\:{the} \\ $$$${number}\:{of}\:{solutions}\:{of}\:{eqn}.\:\left({I}\right). \\ $$$$ \\ $$$${example}:\:{n}=\mathrm{6},\:{m}=\mathrm{13} \\ $$$${GF}={x}\left(\mathrm{1}−{x}^{\mathrm{9}} \right)\left(\mathrm{1}−\mathrm{5}{x}^{\mathrm{10}} +...\right)\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{5}} ^{{k}+\mathrm{5}} {x}^{{k}} \\ $$$$={x}\left(\mathrm{1}−{x}^{\mathrm{9}} −\mathrm{5}{x}^{\mathrm{10}} +...\right)\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{5}} ^{{k}+\mathrm{5}} {x}^{{k}} \\ $$$${coefficient}\:{of}\:{x}^{\mathrm{13}} \:{term}\:{is} \\ $$$$\left({for}\:{k}=\mathrm{12},\:\mathrm{3},\:\mathrm{2}\right) \\ $$$${C}_{\mathrm{5}} ^{\mathrm{17}} −{C}_{\mathrm{5}} ^{\mathrm{8}} −\mathrm{5}{C}_{\mathrm{5}} ^{\mathrm{7}} =\mathrm{6027} \\ $$

Commented by PRITHWISH SEN 2 last updated on 11/Jul/20

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

Answered by prakash jain last updated on 11/Jul/20

xxxxxxxxxxxxxxxxxx (18x) Replace any 5 of last 17 x by a bar count the number of xs between to bars to get a digit. Now this will include cases where where all 5 bars are included in 8 or less continous xs. Meaning digits has to be ≤9. # of ways to place 5 bars=^(17) C_5 one digits is 10=6×^5 C_3 +^7 C_4 +^6 C_4 =110 one digits is 11=5×^4 C_3 +^6 C_4 +^5 C_4 =55 one digits is 12=4×^3 C_3 +^5 C_4 +^4 C_4 =10 one digits is 13=1 valid outcomes =^(17) C_5 −161=6188−161=6027 count the number of x between 2 bars to get the digit. x∣10x∣5x to x6x∣10x xx∣xxx∣xxx∣∣xxx∣xx 233032 xx∣∣∣∣∣∣xxxxxxxxxxx

$${xxxxxxxxxxxxxxxxxx}\:\left(\mathrm{18}{x}\right) \\ $$$$\mathrm{Replace}\:\mathrm{any}\:\mathrm{5}\:\mathrm{of}\:\mathrm{last}\:\mathrm{17}\:{x}\:\mathrm{by}\:\mathrm{a}\:\mathrm{bar} \\ $$$$\mathrm{count}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:{xs}\:\mathrm{between} \\ $$$$\mathrm{to}\:\mathrm{bars}\:\mathrm{to}\:\mathrm{get}\:\mathrm{a}\:\mathrm{digit}. \\ $$$$\mathrm{Now}\:\mathrm{this}\:\mathrm{will}\:\mathrm{include}\:\mathrm{cases}\:\mathrm{where} \\ $$$$\mathrm{where}\:\mathrm{all}\:\mathrm{5}\:\mathrm{bars}\:\mathrm{are}\:\mathrm{included}\:\mathrm{in} \\ $$$$\mathrm{8}\:\mathrm{or}\:\mathrm{less}\:\mathrm{continous}\:{xs}.\:\mathrm{Meaning}\:\mathrm{digits} \\ $$$$\mathrm{has}\:\mathrm{to}\:\mathrm{be}\:\leqslant\mathrm{9}. \\ $$$$#\:\mathrm{of}\:\mathrm{ways}\:\mathrm{to}\:\mathrm{place}\:\mathrm{5}\:\mathrm{bars}=\:^{\mathrm{17}} {C}_{\mathrm{5}} \\ $$$$\mathrm{one}\:\mathrm{digits}\:\mathrm{is}\:\mathrm{10}=\mathrm{6}×\:^{\mathrm{5}} {C}_{\mathrm{3}} +\:^{\mathrm{7}} {C}_{\mathrm{4}} +\:^{\mathrm{6}} {C}_{\mathrm{4}} =\mathrm{110} \\ $$$$\mathrm{one}\:\mathrm{digits}\:\mathrm{is}\:\mathrm{11}=\mathrm{5}×\:^{\mathrm{4}} {C}_{\mathrm{3}} +\:^{\mathrm{6}} {C}_{\mathrm{4}} +\:^{\mathrm{5}} {C}_{\mathrm{4}} =\mathrm{55} \\ $$$$\mathrm{one}\:\mathrm{digits}\:\mathrm{is}\:\mathrm{12}=\mathrm{4}×\:^{\mathrm{3}} {C}_{\mathrm{3}} +\:^{\mathrm{5}} {C}_{\mathrm{4}} +^{\mathrm{4}} {C}_{\mathrm{4}} =\mathrm{10} \\ $$$$\mathrm{one}\:\mathrm{digits}\:\mathrm{is}\:\mathrm{13}=\mathrm{1} \\ $$$$\mathrm{valid}\:\mathrm{outcomes} \\ $$$$=^{\mathrm{17}} \mathrm{C}_{\mathrm{5}} −\mathrm{161}=\mathrm{6188}−\mathrm{161}=\mathrm{6027} \\ $$$$\mathrm{count}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:{x}\:\mathrm{between} \\ $$$$\mathrm{2}\:\mathrm{bars}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{digit}. \\ $$$${x}\mid\mathrm{10}{x}\mid\mathrm{5}{x}\:\mathrm{to}\:{x}\mathrm{6}{x}\mid\mathrm{10}{x} \\ $$$${xx}\mid{xxx}\mid{xxx}\mid\mid{xxx}\mid{xx} \\ $$$$\mathrm{233032} \\ $$$${xx}\mid\mid\mid\mid\mid\mid{xxxxxxxxxxx} \\ $$

Commented by Rasheed.Sindhi last updated on 11/Jul/20

Sir really useful method! I remember now that long ago I had learnt it from you but I forgot!

$${Sir}\:{really}\:{useful}\:{method}!\:{I} \\ $$$${remember}\:{now}\:{that}\:{long}\:{ago}\:{I} \\ $$$${had}\:{learnt}\:{it}\:{from}\:{you}\:{but}\:{I} \\ $$$${forgot}! \\ $$

Commented by prakash jain last updated on 11/Jul/20

This method is commonly known as stars and bars. Very useful for dividing n items in m boxes. The above is a special cases where first box is not empty and other boxes can be empty but cannot contain more than 9 items.

$$\mathrm{This}\:\mathrm{method}\:\mathrm{is}\:\mathrm{commonly}\:\mathrm{known} \\ $$$$\mathrm{as}\:\mathrm{stars}\:\mathrm{and}\:\mathrm{bars}. \\ $$$$\mathrm{Very}\:\mathrm{useful}\:\mathrm{for}\:\mathrm{dividing}\:{n}\:\mathrm{items} \\ $$$$\mathrm{in}\:{m}\:\mathrm{boxes}. \\ $$$$\mathrm{The}\:\mathrm{above}\:\mathrm{is}\:\mathrm{a}\:\mathrm{special}\:\mathrm{cases}\:\mathrm{where} \\ $$$$\mathrm{first}\:\mathrm{box}\:\mathrm{is}\:\mathrm{not}\:\mathrm{empty}\:\mathrm{and}\:\mathrm{other} \\ $$$$\mathrm{boxes}\:\mathrm{can}\:\mathrm{be}\:\mathrm{empty}\:\mathrm{but}\:\mathrm{cannot} \\ $$$$\mathrm{contain}\:\mathrm{more}\:\mathrm{than}\:\mathrm{9}\:\mathrm{items}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com