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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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