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Question Number 99175 by Rio Michael last updated on 19/Jun/20

$$\mathrm{A}\mathrm{man}\mathrm{and}\mathrm{a}\mathrm{woman}\mathrm{have}3\mathrm{boys}\mathrm{and}3\mathrm{girls}.$$$$\left(\mathrm{i}\right)\mathrm{in}\mathrm{how}\mathrm{many}\mathrm{ways}\mathrm{will}\mathrm{they}\mathrm{sit}\mathrm{in}\mathrm{a}\mathrm{row}\mathrm{such}\mathrm{that}$$$$\mathrm{the}3\mathrm{boys}\mathrm{and}3\mathrm{girls}\mathrm{are}\mathrm{inbetween}\mathrm{the}\mathrm{man}\mathrm{and}\mathrm{woman}.$$$$\left(\mathrm{ii}\right)\mathrm{Say}\mathrm{the}\mathrm{man}\mathrm{decides}\mathrm{of}\mathrm{the}3\mathrm{boys}\mathrm{and}3\mathrm{girls}\mathrm{he}\mathrm{has}2$$$$\mathrm{of}\mathrm{the}\mathrm{kids}\mathrm{should}\mathrm{help}\mathrm{him}\mathrm{out}\mathrm{in}\mathrm{a}\mathrm{project},$$$$\mathrm{in}\mathrm{how}\mathrm{many}\mathrm{ways}\mathrm{can}\mathrm{this}\mathrm{be}\mathrm{done},\mathrm{if}\mathrm{heyoungest}\mathrm{boy}\mathrm{and}\mathrm{oldest}$$$$\mathrm{girl}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{join}.$$

Answered by john santu last updated on 19/Jun/20

$$\left(1\right)2!\times 3!\times 2!\times 3!={(2!\times 3!)}^2=144$$

Commented by Rio Michael last updated on 19/Jun/20

$$\mathrm{thanks}\mathrm{sir},\mathrm{you}\mathrm{got}\mathrm{any}\mathrm{explanation}?$$

Commented by mr W last updated on 19/Jun/20

$$ithink$$$$2\times 6!=1440$$

Commented by john santu last updated on 19/Jun/20

$$\mathrm{M}\stackrel{{\mathrm{B}}_1}{-}\stackrel{{\mathrm{B}}_2}{-}\stackrel{{\mathrm{B}}_3}{-}\mathrm{W}\stackrel{{\mathrm{G}}_1}{-}\stackrel{{\mathrm{G}}_2}{-}\stackrel{{\mathrm{G}}_3}{-}=1\times 3!\times 3!$$$$\mathrm{W}\stackrel{{\mathrm{B}}_1}{-}\stackrel{{\mathrm{B}}_2}{-}\stackrel{{\mathrm{B}}_3}{-}\mathrm{M}\stackrel{{\mathrm{G}}_1}{-}\stackrel{{\mathrm{G}}_2}{-}\stackrel{{\mathrm{G}}_3}{-}=1\times 3!\times 3!$$$$\mathrm{M}\stackrel{{\mathrm{G}}_1}{-}\stackrel{{\mathrm{G}}_2}{-}\stackrel{{\mathrm{G}}_3}{-}\mathrm{W}\stackrel{{\mathrm{B}}_1}{-}\stackrel{{\mathrm{B}}_2}{-}\stackrel{{\mathrm{B}}_3}{-}=1\times 3!\times 3!$$$$\mathrm{W}\stackrel{{\mathrm{G}}_1}{-}\stackrel{{\mathrm{G}}_2}{-}\stackrel{{\mathrm{G}}_3}{-}\mathrm{M}\stackrel{{\mathrm{B}}_1}{-}\stackrel{{\mathrm{B}}_2}{-}\stackrel{{\mathrm{B}}_3}{-}=1\times 3!\times 3!$$

Commented by john santu last updated on 19/Jun/20

$$\mathrm{no}\mathrm{sir}$$

Commented by mr W last updated on 19/Jun/20

$$iunderstandthequestionsimplyas$$$$P_1{C}_1{C}_2{C}_3{C}_4{C}_5{C}_6{P}_2$$$$P=parents$$$$C=children$$

Commented by john santu last updated on 19/Jun/20

$$\mathrm{o},\mathrm{i}\mathrm{understand}3\mathrm{boys}\mathrm{or}3\mathrm{girls}\mathrm{between}\mathrm{a}\mathrm{man}\mathrm{\&}\mathrm{a}\mathrm{woman}$$

Answered by john santu last updated on 19/Jun/20

$$\left(2\right)\mathrm{C}(4,2)=\frac{4!}{2!.2!}=12$$

Commented by Rio Michael last updated on 19/Jun/20

$$\mathrm{thanks}{\mathrm{you}}^{\prime }\mathrm{all}$$

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