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Question Number 164560 by PRITHWISH SEN 2 last updated on 18/Jan/22

6 boys and 6 girls go to an exhibition and the cost of ticket is Rs 10.Each girl has a 10 rupees note while each boy has a 20 rupees note. They stand in a queue at the counter and the cashier does not have any money at the begining , then the number of ways of arranging the boys and girls so that no one waits for a change is A) 132 B) 264 C) 132(720)^2 D)264(720)^2

$$\mathrm{6}\:\mathrm{boys}\:\mathrm{and}\:\mathrm{6}\:\mathrm{girls}\:\mathrm{go}\:\mathrm{to}\:\mathrm{an}\:\mathrm{exhibition}\:\mathrm{and}\:\mathrm{the}\:\mathrm{cost} \\ $$$$\mathrm{of}\:\mathrm{ticket}\:\mathrm{is}\:\mathrm{Rs}\:\mathrm{10}.\mathrm{Each}\:\mathrm{girl}\:\mathrm{has}\:\mathrm{a}\:\mathrm{10}\:\mathrm{rupees}\:\mathrm{note} \\ $$$$\mathrm{while}\:\mathrm{each}\:\mathrm{boy}\:\mathrm{has}\:\mathrm{a}\:\mathrm{20}\:\mathrm{rupees}\:\mathrm{note}.\:\mathrm{They}\:\mathrm{stand}\: \\ $$$$\mathrm{in}\:\mathrm{a}\:\mathrm{queue}\:\mathrm{at}\:\mathrm{the}\:\mathrm{counter}\:\mathrm{and}\:\mathrm{the}\:\mathrm{cashier}\:\mathrm{does} \\ $$$$\mathrm{not}\:\mathrm{have}\:\mathrm{any}\:\mathrm{money}\:\mathrm{at}\:\mathrm{the}\:\mathrm{begining}\:,\:\mathrm{then}\:\mathrm{the}\: \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{of}\:\mathrm{arranging}\:\mathrm{the}\:\mathrm{boys}\:\mathrm{and}\:\mathrm{girls} \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{no}\:\mathrm{one}\:\mathrm{waits}\:\mathrm{for}\:\mathrm{a}\:\mathrm{change}\:\mathrm{is} \\ $$$$\left.\mathrm{A}\left.\right)\left.\:\left.\mathrm{132}\:\:\:\:\:\mathrm{B}\right)\:\:\:\:\mathrm{264}\:\mathrm{C}\right)\:\:\mathrm{132}\left(\mathrm{720}\right)^{\mathrm{2}} \:\:\:\:\mathrm{D}\right)\mathrm{264}\left(\mathrm{720}\right)^{\mathrm{2}} \\ $$

Commented by mr W last updated on 12/Mar/23

general solution see Q189049

$${general}\:{solution}\:{see}\:{Q}\mathrm{189049} \\ $$

Answered by nikif99 last updated on 19/Jan/22

C) 132(720)^2 =68 428 800

$$\left.{C}\right)\:\mathrm{132}\left(\mathrm{720}\right)^{\mathrm{2}} =\mathrm{68}\:\mathrm{428}\:\mathrm{800} \\ $$

Commented by PRITHWISH SEN 2 last updated on 20/Jan/22

Thank you sir. But workings ?

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{But}\:\mathrm{workings}\:? \\ $$

Commented by nikif99 last updated on 20/Jan/22

Let′s ignore boys. 6 girls can stand in a queue in 6! ways. Now we ignore girls. 6 boys can stand in a queue in 6! ways. Merging these queues in any way produces (6!)^2 =518400 ways for reaching the cashier. Now the problem is “in how many ways can we merge these two lists with the restriction ′number of girls passed >= number of boys passed′”. I still look for this answer. All queues must start with girl and end with boy. Using graphic design I estimate this to 132. I suppose number of accepted mergins (132) is combinations of 12 over 6 (C_6 ^(12) ) minus 792 which represent queues not fullfilling the restriction (need to explore how to compute it). If someone else could advance to a solution I would be happy to read.

$${Let}'{s}\:{ignore}\:{boys}.\:\mathrm{6}\:{girls}\:{can}\:{stand}\:{in} \\ $$$${a}\:{queue}\:{in}\:\mathrm{6}!\:{ways}.\:{Now}\:{we}\:{ignore}\:{girls}. \\ $$$$\mathrm{6}\:{boys}\:{can}\:{stand}\:{in}\:{a}\:{queue}\:{in}\:\mathrm{6}!\:{ways}. \\ $$$${Merging}\:{these}\:{queues}\:{in}\:{any}\:{way} \\ $$$${produces}\:\left(\mathrm{6}!\right)^{\mathrm{2}} =\mathrm{518400}\:{ways}\:{for} \\ $$$${reaching}\:{the}\:{cashier}.\:{Now}\:{the}\:{problem} \\ $$$${is}\:``{in}\:{how}\:{many}\:{ways}\:{can}\:{we}\:{merge} \\ $$$${these}\:{two}\:{lists}\:{with}\:{the}\:{restriction} \\ $$$$'{number}\:{of}\:{girls}\:{passed}\:>=\:{number} \\ $$$${of}\:{boys}\:{passed}'''. \\ $$$${I}\:{still}\:{look}\:{for}\:{this}\:{answer}.\:{All}\:{queues} \\ $$$${must}\:{start}\:{with}\:{girl}\:{and}\:{end}\:{with}\:{boy}. \\ $$$${Using}\:{graphic}\:{design}\:{I}\:{estimate}\:{this} \\ $$$${to}\:\mathrm{132}.\:{I}\:{suppose}\:{number}\:{of}\:{accepted} \\ $$$${mergins}\:\left(\mathrm{132}\right)\:{is}\:{combinations}\:{of}\:\mathrm{12} \\ $$$${over}\:\mathrm{6}\:\left({C}_{\mathrm{6}} ^{\mathrm{12}} \right)\:{minus}\:\mathrm{792}\:{which}\:{represent} \\ $$$${queues}\:{not}\:{fullfilling}\:{the}\:{restriction} \\ $$$$\left({need}\:{to}\:{explore}\:{how}\:{to}\:{compute}\:{it}\right). \\ $$$${If}\:{someone}\:{else}\:{could}\:{advance}\:{to}\:{a} \\ $$$${solution}\:{I}\:{would}\:{be}\:{happy}\:{to}\:{read}. \\ $$

Answered by mr W last updated on 24/Jan/22

◂ GB_1 GB_2 GB_3 GB_4 GB_5 GB_6 G=positions of the girls B_i =possible positions of boys a=number of boys at position B_1 b=number of boys at position B_2 ... f=number of boys at position B_6 a+b+c+d+e+f=6 0≤a≤1 0≤b≤2−a 0≤c≤3−a−b 0≤d≤4−a−b−c 0≤e≤5−a−b−c−d 0≤f≤6−a−b−c−d−e number of ways to place the 6 boys into the proper positions is n=Σ_(a=0) ^1 Σ_(b=0) ^(2−a) Σ_(c=0) ^(3−a−b) Σ_(d=0) ^(4−a−b−c) Σ_(e=0) ^(5−a−b−c−d) Σ_(f=0) ^(6−a−b−c−d−e) if (a+b+c+d+e+f=6) we get n=132. in each of these arrangements we have 6!=720 ways to arrange the boys and 6!=720 ways to arrange the girls. so the total number of ways to queue the boys and girls properly is 132×(720)^2 .

$$\blacktriangleleft\:{GB}_{\mathrm{1}} {GB}_{\mathrm{2}} {GB}_{\mathrm{3}} {GB}_{\mathrm{4}} {GB}_{\mathrm{5}} {GB}_{\mathrm{6}} \\ $$$${G}={positions}\:{of}\:{the}\:{girls} \\ $$$${B}_{{i}} ={possible}\:{positions}\:{of}\:{boys} \\ $$$${a}={number}\:{of}\:{boys}\:{at}\:{position}\:{B}_{\mathrm{1}} \\ $$$${b}={number}\:{of}\:{boys}\:{at}\:{position}\:{B}_{\mathrm{2}} \\ $$$$... \\ $$$${f}={number}\:{of}\:{boys}\:{at}\:{position}\:{B}_{\mathrm{6}} \\ $$$${a}+{b}+{c}+{d}+{e}+{f}=\mathrm{6} \\ $$$$\mathrm{0}\leqslant{a}\leqslant\mathrm{1} \\ $$$$\mathrm{0}\leqslant{b}\leqslant\mathrm{2}−{a} \\ $$$$\mathrm{0}\leqslant{c}\leqslant\mathrm{3}−{a}−{b} \\ $$$$\mathrm{0}\leqslant{d}\leqslant\mathrm{4}−{a}−{b}−{c} \\ $$$$\mathrm{0}\leqslant{e}\leqslant\mathrm{5}−{a}−{b}−{c}−{d} \\ $$$$\mathrm{0}\leqslant{f}\leqslant\mathrm{6}−{a}−{b}−{c}−{d}−{e} \\ $$$${number}\:{of}\:{ways}\:{to}\:{place}\:{the}\:\mathrm{6}\:{boys} \\ $$$${into}\:{the}\:{proper}\:{positions}\:{is} \\ $$$${n}=\underset{{a}=\mathrm{0}} {\overset{\mathrm{1}} {\sum}}\:\underset{{b}=\mathrm{0}} {\overset{\mathrm{2}−{a}} {\sum}}\:\underset{{c}=\mathrm{0}} {\overset{\mathrm{3}−{a}−{b}} {\sum}}\:\underset{{d}=\mathrm{0}} {\overset{\mathrm{4}−{a}−{b}−{c}} {\sum}}\:\underset{{e}=\mathrm{0}} {\overset{\mathrm{5}−{a}−{b}−{c}−{d}} {\sum}}\:\underset{{f}=\mathrm{0}} {\overset{\mathrm{6}−{a}−{b}−{c}−{d}−{e}} {\sum}}{if}\:\left({a}+{b}+{c}+{d}+{e}+{f}=\mathrm{6}\right) \\ $$$${we}\:{get}\:{n}=\mathrm{132}. \\ $$$${in}\:{each}\:{of}\:{these}\:{arrangements}\:{we} \\ $$$${have}\:\mathrm{6}!=\mathrm{720}\:{ways}\:{to}\:{arrange}\:{the}\:{boys} \\ $$$${and}\:\mathrm{6}!=\mathrm{720}\:{ways}\:{to}\:{arrange}\:{the}\:{girls}. \\ $$$${so}\:{the}\:{total}\:{number}\:{of}\:{ways}\:{to} \\ $$$${queue}\:{the}\:{boys}\:{and}\:{girls}\:{properly}\:{is}\: \\ $$$$\mathrm{132}×\left(\mathrm{720}\right)^{\mathrm{2}} . \\ $$

Commented by mr W last updated on 25/Jan/22

how to calculate n=Σ_(a=0) ^1 Σ_(b=0) ^(2−a) Σ_(c=0) ^(3−a−b) Σ_(d=0) ^(4−a−b−c) Σ_(e=0) ^(5−a−b−c−d) Σ_(f=0) ^(6−a−b−c−d−e) if (a+b+c+d+e+f=6) we can get the value of n in following way, e.g. using wolfram alpha: n_1 =Σ_(a=0) ^1 Σ_(b=0) ^(2−a) Σ_(c=0) ^(3−a−b) Σ_(d=0) ^(4−a−b−c) Σ_(e=0) ^(5−a−b−c−d) Σ_(f=0) ^(6−a−b−c−d−e) (1) n_2 =Σ_(a=0) ^1 Σ_(b=0) ^(2−a) Σ_(c=0) ^(3−a−b) Σ_(d=0) ^(4−a−b−c) Σ_(e=0) ^(5−a−b−c−d) Σ_(f=0) ^(6−a−b−c−d−e) ∣sign (a+b+c+d+e+f−6)∣ n_1 =429 n_2 =297 n=n_1 −n_2 =429−297=132

$${how}\:{to}\:{calculate} \\ $$$${n}=\underset{{a}=\mathrm{0}} {\overset{\mathrm{1}} {\sum}}\:\underset{{b}=\mathrm{0}} {\overset{\mathrm{2}−{a}} {\sum}}\:\underset{{c}=\mathrm{0}} {\overset{\mathrm{3}−{a}−{b}} {\sum}}\:\underset{{d}=\mathrm{0}} {\overset{\mathrm{4}−{a}−{b}−{c}} {\sum}}\:\underset{{e}=\mathrm{0}} {\overset{\mathrm{5}−{a}−{b}−{c}−{d}} {\sum}}\:\underset{{f}=\mathrm{0}} {\overset{\mathrm{6}−{a}−{b}−{c}−{d}−{e}} {\sum}}{if}\:\left({a}+{b}+{c}+{d}+{e}+{f}=\mathrm{6}\right) \\ $$$${we}\:{can}\:{get}\:{the}\:{value}\:{of}\:{n}\:{in}\:{following} \\ $$$${way},\:{e}.{g}.\:{using}\:{wolfram}\:{alpha}: \\ $$$${n}_{\mathrm{1}} =\underset{{a}=\mathrm{0}} {\overset{\mathrm{1}} {\sum}}\:\underset{{b}=\mathrm{0}} {\overset{\mathrm{2}−{a}} {\sum}}\:\underset{{c}=\mathrm{0}} {\overset{\mathrm{3}−{a}−{b}} {\sum}}\:\underset{{d}=\mathrm{0}} {\overset{\mathrm{4}−{a}−{b}−{c}} {\sum}}\:\underset{{e}=\mathrm{0}} {\overset{\mathrm{5}−{a}−{b}−{c}−{d}} {\sum}}\:\underset{{f}=\mathrm{0}} {\overset{\mathrm{6}−{a}−{b}−{c}−{d}−{e}} {\sum}}\left(\mathrm{1}\right) \\ $$$${n}_{\mathrm{2}} =\underset{{a}=\mathrm{0}} {\overset{\mathrm{1}} {\sum}}\:\underset{{b}=\mathrm{0}} {\overset{\mathrm{2}−{a}} {\sum}}\:\underset{{c}=\mathrm{0}} {\overset{\mathrm{3}−{a}−{b}} {\sum}}\:\underset{{d}=\mathrm{0}} {\overset{\mathrm{4}−{a}−{b}−{c}} {\sum}}\:\underset{{e}=\mathrm{0}} {\overset{\mathrm{5}−{a}−{b}−{c}−{d}} {\sum}}\:\underset{{f}=\mathrm{0}} {\overset{\mathrm{6}−{a}−{b}−{c}−{d}−{e}} {\sum}}\mid{sign}\:\left({a}+{b}+{c}+{d}+{e}+{f}−\mathrm{6}\right)\mid \\ $$$${n}_{\mathrm{1}} =\mathrm{429} \\ $$$${n}_{\mathrm{2}} =\mathrm{297} \\ $$$${n}={n}_{\mathrm{1}} −{n}_{\mathrm{2}} =\mathrm{429}−\mathrm{297}=\mathrm{132} \\ $$

Commented by mr W last updated on 25/Jan/22

Commented by mr W last updated on 25/Jan/22

Commented by nikif99 last updated on 30/Jan/22

A wonderfull solution to a complicated problem. Thank you Sir W for your contribution.

$${A}\:{wonderfull}\:{solution}\:{to}\:{a}\:{complicated} \\ $$$${problem}.\:{Thank}\:{you}\:{Sir}\:{W}\:{for}\:{your} \\ $$$${contribution}. \\ $$

Commented by Ar Brandon last updated on 30/Jan/22

Q161675 , Sir ��

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