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Question Number 125191 by Don08q last updated on 08/Dec/20
In how many ways can 3 men and 5 women be seated in a row if 3 specific women cannot sit next to each other? My solution: arrangements when the 3 women cannot sit next to eachother = 5! ×^6 P_3 = 14,400 Solution in book: arrangements with no restrictions = 8! arrangements with the 3 women seated next to eachother = 3! × 6! arrangements when the 3 women cannot sit next to eachother = 8! − (3! × 6!) = 36,000 I′m confused. Can I please get some explanation why my solution is wrong?
$$\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{3}\:\mathrm{men}\:\mathrm{and}\:\mathrm{5} \\ $$$$\mathrm{women}\:\mathrm{be}\:\mathrm{seated}\:\mathrm{in}\:\mathrm{a}\:\mathrm{row}\:\mathrm{if}\:\mathrm{3}\:\mathrm{specific} \\ $$$$\mathrm{women}\:{cannot}\:\mathrm{sit}\:\mathrm{next}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other}? \\ $$$$ \\ $$$$\mathrm{My}\:\mathrm{solution}: \\ $$$$\mathrm{arrangements}\:\mathrm{when}\:\mathrm{the}\:\mathrm{3}\:\mathrm{women} \\ $$$${cannot}\:\mathrm{sit}\:\mathrm{next}\:\mathrm{to}\:\mathrm{eachother}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{5}!\:×\:^{\mathrm{6}} {P}_{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{14},\mathrm{400} \\ $$$$ \\ $$$$\mathrm{Solution}\:\mathrm{in}\:\mathrm{book}: \\ $$$$\mathrm{arrangements}\:\mathrm{with}\:\mathrm{no}\:\mathrm{restrictions}\:=\:\mathrm{8}! \\ $$$$ \\ $$$$\mathrm{arrangements}\:\mathrm{with}\:\mathrm{the}\:\mathrm{3}\:\mathrm{women} \\ $$$$\mathrm{seated}\:\mathrm{next}\:\mathrm{to}\:\mathrm{eachother}\:\:=\:\:\mathrm{3}!\:×\:\mathrm{6}! \\ $$$$ \\ $$$$\mathrm{arrangements}\:\mathrm{when}\:\mathrm{the}\:\mathrm{3}\:\mathrm{women} \\ $$$${cannot}\:\mathrm{sit}\:\mathrm{next}\:\mathrm{to}\:\mathrm{eachother}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{8}!\:−\:\left(\mathrm{3}!\:×\:\mathrm{6}!\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{36},\mathrm{000} \\ $$$$ \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{confused}.\:\mathrm{Can}\:\mathrm{I}\:\mathrm{please}\:\mathrm{get}\:\mathrm{some} \\ $$$$\mathrm{explanation}\:\mathrm{why}\:\mathrm{my}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{wrong}? \\ $$
Commented by liberty last updated on 09/Dec/20
(•) − W_1 −W_2 −W_3 − treat 3 men and 2 remaining women as identical ′x′. Since W_1 ,W_2 and W_3 no adjacent ⇒ −W_1 −^x W_2 −^x W_3 − the 3 remaining ′x′s can be put the 4 places arbitrarely in (((3+4−1)),(( 3)) ) = ((6),(3) ) = 20 the number of way arrangement is given by 20×3!×5! = 120×120=14400
$$\left(\bullet\right)\:−\:{W}_{\mathrm{1}} −{W}_{\mathrm{2}} −{W}_{\mathrm{3}} − \\ $$$${treat}\:\mathrm{3}\:{men}\:{and}\:\mathrm{2}\:{remaining}\:{women}\:{as} \\ $$$${identical}\:'{x}'.\:{Since}\:{W}_{\mathrm{1}} ,{W}_{\mathrm{2}} \:{and}\:{W}_{\mathrm{3}} \:{no}\: \\ $$$${adjacent}\:\Rightarrow\:−{W}_{\mathrm{1}} \overset{{x}} {−}{W}_{\mathrm{2}} \overset{{x}} {−}{W}_{\mathrm{3}} − \\ $$$${the}\:\mathrm{3}\:{remaining}\:'{x}'{s}\:{can}\:{be}\:{put}\:{the}\:\mathrm{4}\:{places} \\ $$$${arbitrarely}\:{in}\:\begin{pmatrix}{\mathrm{3}+\mathrm{4}−\mathrm{1}}\\{\:\:\:\:\:\:\:\mathrm{3}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{6}}\\{\mathrm{3}}\end{pmatrix}\:=\:\mathrm{20} \\ $$$${the}\:{number}\:{of}\:{way}\:{arrangement}\:{is} \\ $$$${given}\:{by}\:\mathrm{20}×\mathrm{3}!×\mathrm{5}!\:=\:\mathrm{120}×\mathrm{120}=\mathrm{14400} \\ $$$$ \\ $$
Commented by mr W last updated on 09/Dec/20
both of you are wrong sirs. you treated the cases that “no two of the three women are next to each other”. but the question askes that “all the three women are not next to each other”. that means according to the question we should only exclude the cases that all the three women are together. but in your solutions you exclude also the cases that two of the three women are together. the solution given in book is correct.
$${both}\:{of}\:{you}\:{are}\:{wrong}\:{sirs}. \\ $$$${you}\:{treated}\:{the}\:{cases}\:{that}\:“{no}\:{two}\:{of} \\ $$$${the}\:{three}\:{women}\:{are}\:{next}\:{to}\:{each} \\ $$$${other}''.\:{but}\:{the}\:{question}\:{askes}\:{that} \\ $$$$“{all}\:{the}\:{three}\:{women}\:{are}\:{not}\:{next} \\ $$$${to}\:{each}\:{other}''.\:{that}\:{means}\:{according} \\ $$$${to}\:{the}\:{question}\:{we}\:{should}\:{only}\:{exclude} \\ $$$${the}\:{cases}\:{that}\:{all}\:{the}\:{three}\:{women} \\ $$$${are}\:{together}.\:{but}\:{in}\:{your}\:{solutions} \\ $$$${you}\:{exclude}\:{also}\:{the}\:{cases}\:{that}\:{two} \\ $$$${of}\:{the}\:{three}\:{women}\:{are}\:{together}. \\ $$$${the}\:{solution}\:{given}\:{in}\:{book}\:{is}\:{correct}. \\ $$
Commented by mr W last updated on 09/Dec/20
again: according to question it is allowed when two of the three women are together. but this is not allowed in your solutions.
$${again}:\:{according}\:{to}\:{question}\:{it}\:{is} \\ $$$${allowed}\:{when}\:{two}\:{of}\:{the}\:{three}\:{women} \\ $$$${are}\:{together}.\:{but}\:{this}\:{is}\:{not}\:{allowed} \\ $$$${in}\:{your}\:{solutions}. \\ $$
Commented by liberty last updated on 11/Dec/20
yeah....i misunderstanding the question.
$${yeah}….{i}\:{misunderstanding}\:{the}\:{question}. \\ $$

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