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100-apples-should-be-packed-in-three-boxes-and-each-box-should-contain-at-least-10-apples-in-how-many-ways-can-this-be-done-

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Question Number 84993 by mr W last updated on 18/Mar/20
100 apples should be packed in three boxes and each box should contain at least 10 apples. in how many ways can this be done?
$$\mathrm{100}\:{apples}\:{should}\:{be}\:{packed}\:{in}\:{three} \\ $$$${boxes}\:{and}\:{each}\:{box}\:{should}\:{contain} \\ $$$${at}\:{least}\:\mathrm{10}\:{apples}.\:{in}\:{how}\:{many}\:{ways} \\ $$$${can}\:{this}\:{be}\:{done}? \\ $$
Commented by john santu last updated on 18/Mar/20
let the box we call it ABC [(10,10,80) (10,11,79) , (10,12,78), ... (10, 80,10) ] × 2
$$\mathrm{let}\:\mathrm{the}\:\mathrm{box}\:\mathrm{we}\:\mathrm{call}\:\mathrm{it}\:\mathrm{ABC} \\ $$$$\left[\left(\mathrm{10},\mathrm{10},\mathrm{80}\right)\:\left(\mathrm{10},\mathrm{11},\mathrm{79}\right)\:,\:\left(\mathrm{10},\mathrm{12},\mathrm{78}\right),\right. \\ $$$$\left….\:\left(\mathrm{10},\:\mathrm{80},\mathrm{10}\right)\:\right]\:×\:\mathrm{2} \\ $$
Commented by john santu last updated on 18/Mar/20
⇒ 71 ×2 = 142 ways
$$\Rightarrow\:\mathrm{71}\:×\mathrm{2}\:=\:\mathrm{142}\:\mathrm{ways} \\ $$
Commented by mr W last updated on 18/Mar/20
you treat the boxes as different? what′s the meaning of ×2 ?
$${you}\:{treat}\:{the}\:{boxes}\:{as}\:{different}?\: \\ $$$${what}'{s}\:{the}\:{meaning}\:{of}\:×\mathrm{2}\:? \\ $$
Commented by john santu last updated on 18/Mar/20
yes. i assume the boxes are different. ×2 is mean has 2 case case 2 (11,10,79)(12,10,78)...(80,10,10)
$$\mathrm{yes}.\:\mathrm{i}\:\mathrm{assume}\:\mathrm{the}\:\mathrm{boxes}\:\mathrm{are}\:\mathrm{different}. \\ $$$$×\mathrm{2}\:\mathrm{is}\:\mathrm{mean}\:\mathrm{has}\:\mathrm{2}\:\mathrm{case} \\ $$$$\mathrm{case}\:\mathrm{2}\: \\ $$$$\left(\mathrm{11},\mathrm{10},\mathrm{79}\right)\left(\mathrm{12},\mathrm{10},\mathrm{78}\right)…\left(\mathrm{80},\mathrm{10},\mathrm{10}\right) \\ $$
Commented by mr W last updated on 18/Mar/20
as example (11,11,78) is not included in your consideration.
$${as}\:{example}\:\left(\mathrm{11},\mathrm{11},\mathrm{78}\right)\:{is}\:{not}\:{included} \\ $$$${in}\:{your}\:{consideration}. \\ $$
Commented by mr W last updated on 19/Mar/20
it′s wrong sir. you didn′t take into account that each box should have at least 10 apples. the formula you applied is for the case that a box may get zero apple. besides in our question there are three boxes, not 4 as you took.
$${it}'{s}\:{wrong}\:{sir}. \\ $$$${you}\:{didn}'{t}\:{take}\:{into}\:{account}\:{that}\:{each} \\ $$$${box}\:{should}\:{have}\:{at}\:{least}\:\mathrm{10}\:{apples}. \\ $$$${the}\:{formula}\:{you}\:{applied}\:{is}\:{for}\:{the}\:{case} \\ $$$${that}\:{a}\:{box}\:{may}\:{get}\:{zero}\:{apple}.\:{besides} \\ $$$${in}\:{our}\:{question}\:{there}\:{are}\:{three}\:{boxes}, \\ $$$${not}\:\mathrm{4}\:{as}\:{you}\:{took}. \\ $$
Commented by Rio Michael last updated on 18/Mar/20
okay i just read on this. thanks mr Wfor this question what i got is ′the number of ways of dividing n identical objects into r groups is ^(n + r−1) C_(r−1) ways. lets apply it to the question i get number of ways =^(100 + 4 − 1) C_(4−1) such that each gets at least 10 =^(103) C_3 = 357760 ways
$$\mathrm{okay}\:\mathrm{i}\:\mathrm{just}\:\mathrm{read}\:\mathrm{on}\:\mathrm{this}.\:\mathrm{thanks}\:\mathrm{mr}\:\mathrm{Wfor}\:\mathrm{this}\:\mathrm{question} \\ $$$$\mathrm{what}\:\mathrm{i}\:\mathrm{got}\:\mathrm{is}\:'\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{of}\:\mathrm{dividing}\:{n}\:\mathrm{identical} \\ $$$$\mathrm{objects}\:\mathrm{into}\:{r}\:\mathrm{groups}\:\mathrm{is}\:\:\:^{{n}\:+\:{r}−\mathrm{1}} {C}_{{r}−\mathrm{1}} \:\mathrm{ways}. \\ $$$$\mathrm{lets}\:\mathrm{apply}\:\mathrm{it}\:\mathrm{to}\:\mathrm{the}\:\mathrm{question} \\ $$$$\mathrm{i}\:\mathrm{get}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:=\:^{\mathrm{100}\:+\:\mathrm{4}\:−\:\mathrm{1}} \mathrm{C}_{\mathrm{4}−\mathrm{1}} \:\mathrm{such}\:\mathrm{that}\:\mathrm{each}\:\mathrm{gets}\:\mathrm{at}\:\mathrm{least}\:\mathrm{10} \\ $$$$\:\:\:=\:^{\mathrm{103}} \mathrm{C}_{\mathrm{3}} \:=\:\mathrm{357760}\:\mathrm{ways} \\ $$
Answered by Rio Michael last updated on 18/Mar/20
this is my approach sir ∗ the first bag can take 10 apples in^(100) C_(10) ways. ∗ the next bag can take 10 apples in^(90) C_(10 ) ways. ∗ the third can take 10 apples in^(80) C_(10) ways ⇒ total number of ways for 90 apples=^(100) C_(10) ×^(90) C_(10) ×^(80) C_(10) × 2 as the number of apples drop down to 90. now we can share the remainder in^(10) C_3 ways ⇒ total number of ways= ^(100) C_(10) ×^(90) C_(10) ×^(80) C_(10) × 2 ×^(10) C_3
$$\mathrm{this}\:\mathrm{is}\:\mathrm{my}\:\mathrm{approach}\:\mathrm{sir} \\ $$$$\ast\:\mathrm{the}\:\mathrm{first}\:\mathrm{bag}\:\mathrm{can}\:\mathrm{take}\:\mathrm{10}\:\mathrm{apples}\:\mathrm{in}\:^{\mathrm{100}} \mathrm{C}_{\mathrm{10}} \:\mathrm{ways}. \\ $$$$\ast\:\mathrm{the}\:\mathrm{next}\:\mathrm{bag}\:\mathrm{can}\:\mathrm{take}\:\mathrm{10}\:\mathrm{apples}\:\mathrm{in}\:^{\mathrm{90}} \mathrm{C}_{\mathrm{10}\:} \:\mathrm{ways}. \\ $$$$\ast\:\mathrm{the}\:\mathrm{third}\:\mathrm{can}\:\mathrm{take}\:\mathrm{10}\:\mathrm{apples}\:\mathrm{in}\:^{\mathrm{80}} \mathrm{C}_{\mathrm{10}} \:\mathrm{ways} \\ $$$$\Rightarrow\:\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{for}\:\mathrm{90}\:\mathrm{apples}=\:^{\mathrm{100}} \mathrm{C}_{\mathrm{10}} \:×\:^{\mathrm{90}} {C}_{\mathrm{10}} \:×\:^{\mathrm{80}} {C}_{\mathrm{10}} \:×\:\mathrm{2}\:\:\:\mathrm{as}\:\mathrm{the}\:\mathrm{number} \\ $$$$\mathrm{of}\:\mathrm{apples}\:\mathrm{drop}\:\mathrm{down}\:\mathrm{to}\:\mathrm{90}. \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{can}\:\mathrm{share}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{in}\:^{\mathrm{10}} \mathrm{C}_{\mathrm{3}} \:\mathrm{ways} \\ $$$$\Rightarrow\:\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}= \\ $$$$\:^{\mathrm{100}} \mathrm{C}_{\mathrm{10}} \:×\:^{\mathrm{90}} {C}_{\mathrm{10}} \:×\:^{\mathrm{80}} {C}_{\mathrm{10}} \:×\:\mathrm{2}\:×\:^{\mathrm{10}} {C}_{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 18/Mar/20
the apples are identical.
$${the}\:{apples}\:{are}\:{identical}. \\ $$
Commented by Rio Michael last updated on 18/Mar/20
ofcourse sir we just select at random
$$\mathrm{ofcourse}\:\mathrm{sir}\:\mathrm{we}\:\mathrm{just}\:\mathrm{select}\:\mathrm{at}\:\mathrm{random} \\ $$
Commented by mr W last updated on 18/Mar/20
but your working says something different: ∗ the first bag can take 10 apples in^(100) C_(10) ways. ⇒this means the apples are distinct. since if the apples are identical and the first bag takes 10 apples, then there is only one possibility, namely it obtains 10 apples. you can not give ihm 10 apples in different ways.
$${but}\:{your}\:{working}\:{says}\:{something} \\ $$$${different}: \\ $$$$\ast\:\mathrm{the}\:\mathrm{first}\:\mathrm{bag}\:\mathrm{can}\:\mathrm{take}\:\mathrm{10}\:\mathrm{apples}\:\mathrm{in}\:^{\mathrm{100}} \mathrm{C}_{\mathrm{10}} \:\mathrm{ways}. \\ $$$$\Rightarrow{this}\:{means}\:{the}\:{apples}\:{are}\:{distinct}. \\ $$$${since}\:{if}\:{the}\:{apples}\:{are}\:{identical}\:{and} \\ $$$${the}\:{first}\:{bag}\:{takes}\:\mathrm{10}\:{apples},\:{then} \\ $$$${there}\:{is}\:{only}\:{one}\:{possibility},\:{namely} \\ $$$${it}\:{obtains}\:\mathrm{10}\:{apples}.\:{you}\:{can}\:{not}\:{give} \\ $$$${ihm}\:\mathrm{10}\:{apples}\:{in}\:{different}\:{ways}. \\ $$
Commented by Rio Michael last updated on 18/Mar/20
you′re right sir
$$\:\: \\ $$$$\mathrm{you}'\mathrm{re}\:\mathrm{right}\:\mathrm{sir}\: \\ $$
Answered by Joel578 last updated on 18/Mar/20
This problem is same with how many integer solutions of x_1 + x_2 + x_3 = 100, 10≤x_i ≤80, i = 1,2,3 and the problem above is same with how many integer solutions of y_1 + y_2 + y_3 = 70, 0≤y_i ≤70 It has (((70 + 3 − 1)),(( 3 − 1)) ) = ((( 72)),(( 2)) ) = ((72(71))/2) = 2556 ways
$$\mathrm{This}\:\mathrm{problem}\:\mathrm{is}\:\mathrm{same}\:\mathrm{with}\:\mathrm{how}\:\mathrm{many}\:\mathrm{integer} \\ $$$$\mathrm{solutions}\:\mathrm{of}\: \\ $$$${x}_{\mathrm{1}} \:+\:{x}_{\mathrm{2}} \:+\:{x}_{\mathrm{3}} \:=\:\mathrm{100},\:\:\:\mathrm{10}\leqslant{x}_{{i}} \leqslant\mathrm{80},\:\:{i}\:=\:\mathrm{1},\mathrm{2},\mathrm{3} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{above}\:\mathrm{is}\:\mathrm{same}\:\mathrm{with}\:\mathrm{how} \\ $$$$\mathrm{many}\:\mathrm{integer}\:\mathrm{solutions}\:\mathrm{of} \\ $$$${y}_{\mathrm{1}} \:+\:{y}_{\mathrm{2}} \:+\:{y}_{\mathrm{3}} \:=\:\mathrm{70},\:\:\:\mathrm{0}\leqslant{y}_{{i}} \leqslant\mathrm{70} \\ $$$$\mathrm{It}\:\mathrm{has}\:\begin{pmatrix}{\mathrm{70}\:+\:\mathrm{3}\:−\:\mathrm{1}}\\{\:\:\:\:\:\:\:\mathrm{3}\:−\:\mathrm{1}}\end{pmatrix}\:=\:\begin{pmatrix}{\:\mathrm{72}}\\{\:\:\mathrm{2}}\end{pmatrix}\:=\:\frac{\mathrm{72}\left(\mathrm{71}\right)}{\mathrm{2}}\:=\:\mathrm{2556}\:\mathrm{ways} \\ $$
Commented by jagoll last updated on 18/Mar/20
how to get (((70+3−1)),(( 3−1)) ) ?
$$\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\begin{pmatrix}{\mathrm{70}+\mathrm{3}−\mathrm{1}}\\{\:\:\:\:\:\:\:\mathrm{3}−\mathrm{1}}\end{pmatrix}\:? \\ $$
Commented by mr W last updated on 18/Mar/20
that′s correct sir! what if the boxes are identical? that means we only divide the apples into three parts, each part with at least 10 apples.
$${that}'{s}\:{correct}\:{sir}! \\ $$$${what}\:{if}\:{the}\:{boxes}\:{are}\:{identical}?\:{that} \\ $$$${means}\:{we}\:{only}\:{divide}\:{the}\:{apples}\:{into} \\ $$$${three}\:{parts},\:{each}\:{part}\:{with}\:{at}\:{least} \\ $$$$\mathrm{10}\:{apples}. \\ $$
Commented by jagoll last updated on 18/Mar/20
if 50 candies are divided to 4 children with each child getting a minimum of 10 candies
$$\mathrm{if}\:\mathrm{50}\:\mathrm{candies}\:\mathrm{are}\:\mathrm{divided}\:\mathrm{to}\:\mathrm{4}\:\mathrm{children} \\ $$$$\mathrm{with}\:\mathrm{each}\:\mathrm{child}\:\mathrm{getting}\:\mathrm{a}\:\mathrm{minimum} \\ $$$$\mathrm{of}\:\mathrm{10}\:\mathrm{candies} \\ $$
Commented by jagoll last updated on 18/Mar/20
(((50+4−1)),(( 4−1)) ) = ( _3 ^(53) ) = ((53.52.51)/(3.2.1)) ?
$$\begin{pmatrix}{\mathrm{50}+\mathrm{4}−\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{4}−\mathrm{1}}\end{pmatrix}\:=\:\left(\underset{\mathrm{3}} {\overset{\mathrm{53}} {\:}}\right)\:=\:\frac{\mathrm{53}.\mathrm{52}.\mathrm{51}}{\mathrm{3}.\mathrm{2}.\mathrm{1}}\:? \\ $$
Commented by mr W last updated on 18/Mar/20
no. each kid gets 9 at first. there are 14 remaining. to share these 14 pieces there are (_(4−1) ^(14−1) )=(_3 ^(13) ) ways. see stars and bars method.
$${no}.\:{each}\:{kid}\:{gets}\:\mathrm{9}\:{at}\:{first}.\:{there}\:{are} \\ $$$$\mathrm{14}\:{remaining}.\:{to}\:{share}\:{these}\:\mathrm{14}\:{pieces} \\ $$$${there}\:{are}\:\left(_{\mathrm{4}−\mathrm{1}} ^{\mathrm{14}−\mathrm{1}} \right)=\left(_{\mathrm{3}} ^{\mathrm{13}} \right)\:{ways}. \\ $$$${see}\:{stars}\:{and}\:{bars}\:{method}. \\ $$
Commented by jagoll last updated on 18/Mar/20
why not ( _(4−1) ^(10−1) ) sir? each kid get minimum 10 candies. there are 10 remaining to share these 10 pieces
$$\mathrm{why}\:\mathrm{not}\:\left(\underset{\mathrm{4}−\mathrm{1}} {\overset{\mathrm{10}−\mathrm{1}} {\:}}\right)\:\mathrm{sir}?\:\mathrm{each}\:\mathrm{kid}\:\mathrm{get}\:\mathrm{minimum} \\ $$$$\mathrm{10}\:\mathrm{candies}.\:\mathrm{there}\:\mathrm{are}\:\mathrm{10}\:\mathrm{remaining}\: \\ $$$$\mathrm{to}\:\mathrm{share}\:\mathrm{these}\:\mathrm{10}\:\mathrm{pieces} \\ $$
Commented by Joel578 last updated on 18/Mar/20
To sir mr W Sir, please elaborate your question again. I didn′t see any differences with the earlier question. Maybe some example will help
$$\mathrm{To}\:\mathrm{sir}\:\mathrm{mr}\:\mathrm{W}\: \\ $$$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{elaborate}\:\mathrm{your}\:\mathrm{question}\:\mathrm{again}. \\ $$$$\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{see}\:\mathrm{any}\:\mathrm{differences}\:\mathrm{with}\:\mathrm{the}\:\mathrm{earlier}\:\mathrm{question}. \\ $$$$\mathrm{Maybe}\:\mathrm{some}\:\mathrm{example}\:\mathrm{will}\:\mathrm{help} \\ $$
Commented by Joel578 last updated on 18/Mar/20
To sir Jagoll Here I send a clear explanation of the formula I used. Hope this will help.
$$\mathrm{To}\:\mathrm{sir}\:\mathrm{Jagoll} \\ $$$$\mathrm{Here}\:\mathrm{I}\:\mathrm{send}\:\mathrm{a}\:\mathrm{clear}\:\mathrm{explanation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{formula} \\ $$$$\mathrm{I}\:\mathrm{used}.\:\mathrm{Hope}\:\mathrm{this}\:\mathrm{will}\:\mathrm{help}. \\ $$
Commented by mr W last updated on 18/Mar/20
if they share the 10 remaining, it means one or two kid may get zero candy, because each of the has already 10 candies before. to share 10 candies among 4 kids, zero is allowed, there are (_(4−1) ^(10+4−1) )=(_3 ^(13) ) ways. formula: to share n objects among r persons, each with ≥1 object, there are (_(r−1) ^(n−1) ) ways. to share n objects among r persons, each with ≥0 object, there are (_(r−1) ^(n+r−1) ) ways.
$${if}\:{they}\:{share}\:{the}\:\mathrm{10}\:{remaining},\:{it} \\ $$$${means}\:{one}\:{or}\:{two}\:{kid}\:{may}\:{get}\:{zero} \\ $$$${candy},\:{because}\:{each}\:{of}\:{the}\:{has}\:{already} \\ $$$$\mathrm{10}\:{candies}\:{before}.\:{to}\:{share}\:\mathrm{10}\:{candies} \\ $$$${among}\:\mathrm{4}\:{kids},\:{zero}\:{is}\:{allowed},\:{there} \\ $$$${are}\:\left(_{\mathrm{4}−\mathrm{1}} ^{\mathrm{10}+\mathrm{4}−\mathrm{1}} \right)=\left(_{\mathrm{3}} ^{\mathrm{13}} \right)\:{ways}. \\ $$$$ \\ $$$${formula}:\:{to}\:{share}\:{n}\:{objects}\:{among} \\ $$$${r}\:{persons},\:{each}\:{with}\:\geqslant\mathrm{1}\:{object},\:{there} \\ $$$${are}\:\left(_{{r}−\mathrm{1}} ^{{n}−\mathrm{1}} \right)\:{ways}. \\ $$$${to}\:{share}\:{n}\:{objects}\:{among} \\ $$$${r}\:{persons},\:{each}\:{with}\:\geqslant\mathrm{0}\:{object},\:{there} \\ $$$${are}\:\left(_{{r}−\mathrm{1}} ^{{n}+{r}−\mathrm{1}} \right)\:{ways}. \\ $$
Commented by john santu last updated on 18/Mar/20
does means that all three boxes considered identical
$$\mathrm{does}\:\mathrm{means}\:\mathrm{that}\:\mathrm{all}\:\mathrm{three}\:\mathrm{boxes}\: \\ $$$$\mathrm{considered}\:\mathrm{identical}\: \\ $$
Commented by Joel578 last updated on 18/Mar/20
Commented by Joel578 last updated on 18/Mar/20
Commented by mr W last updated on 18/Mar/20
to joel sir: when 100 apples are divided into 3 parts, it can be for example 20, 30, 50 this is only one possibility. but when these apples are put into 3 distinct boxes, there are 6 different ways, e.g. A=20, B=30, C=50 A=20, B=50, C=30 A=30, B=20, C=50 A=30, B=50, C=20 A=50, B=20, C=30 A=50, B=30, C=20 this is what we had till now. what i ask now is how many ways are there to divide 100 apples into 3 parts
$${to}\:{joel}\:{sir}: \\ $$$${when}\:\mathrm{100}\:{apples}\:{are}\:{divided}\:{into}\:\mathrm{3} \\ $$$${parts},\:{it}\:{can}\:{be}\:{for}\:{example} \\ $$$$\mathrm{20},\:\mathrm{30},\:\mathrm{50} \\ $$$${this}\:{is}\:{only}\:{one}\:{possibility}.\:{but}\:{when} \\ $$$${these}\:{apples}\:{are}\:{put}\:{into}\:\mathrm{3}\:{distinct}\:{boxes}, \\ $$$${there}\:{are}\:\mathrm{6}\:{different}\:{ways},\:{e}.{g}. \\ $$$${A}=\mathrm{20},\:{B}=\mathrm{30},\:{C}=\mathrm{50} \\ $$$${A}=\mathrm{20},\:{B}=\mathrm{50},\:{C}=\mathrm{30} \\ $$$${A}=\mathrm{30},\:{B}=\mathrm{20},\:{C}=\mathrm{50} \\ $$$${A}=\mathrm{30},\:{B}=\mathrm{50},\:{C}=\mathrm{20} \\ $$$${A}=\mathrm{50},\:{B}=\mathrm{20},\:{C}=\mathrm{30} \\ $$$${A}=\mathrm{50},\:{B}=\mathrm{30},\:{C}=\mathrm{20} \\ $$$${this}\:{is}\:{what}\:{we}\:{had}\:{till}\:{now}. \\ $$$${what}\:{i}\:{ask}\:{now}\:{is}\:{how}\:{many}\:{ways} \\ $$$${are}\:{there}\:{to}\:{divide}\:\mathrm{100}\:{apples}\:{into}\:\mathrm{3} \\ $$$${parts} \\ $$
Commented by jagoll last updated on 19/Mar/20
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by jagoll last updated on 19/Mar/20
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$

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