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Question Number 34930 by NECx last updated on 13/May/18

$$Acommitteeof4istobeformed$$$$from4principals$$$$and5students.Inhowmanyways$$$$canthisbedoneifaparticular$$$$studentandaprincipalmustbe$$$$inthecommittee.$$$$$$

Commented by Rasheed.Sindhi last updated on 13/May/18

$${}^{\bullet }\mathrm{A}\mathrm{particular}\mathrm{principal}\mathrm{can}\mathrm{be}\mathrm{seleted}$$$$\mathrm{in}1\mathrm{way}.$$$${}^{\bullet }\mathrm{A}\mathrm{particular}\mathrm{student}\mathrm{can}\mathrm{be}\mathrm{seleted}$$$$\mathrm{in}1\mathrm{way}.$$$$(\mathrm{Now}\mathrm{there}\mathrm{remain}4+5-2=7\mathrm{persons})$$$${}^{\bullet }\mathrm{So}\mathrm{third}\mathrm{member}\mathrm{may}\mathrm{be}\mathrm{selected}\mathrm{in}$$$$7\mathrm{ways}.$$$$(\mathrm{Now}\mathrm{there}\mathrm{remain}7-1=6\mathrm{persons}.$$$${}^{\bullet }\mathrm{The}\mathrm{fourth}\mathrm{member}\mathrm{can}\mathrm{be}\mathrm{selected}$$$$\mathrm{in}6\mathrm{ways}$$$$\mathrm{Total}\mathrm{ways}1\times 1\times 7\times 6=42\mathrm{ways}.$$

Commented by NECx last updated on 13/May/18

$$sinceastudentandaprincipalmust$$$$beincludedthenwehave3principals$$$$lefttochoose1fromand4students$$$$lefttochoose1.$$$$={\stackrel{3}{C}}_1\times {\stackrel{4}{C}}_1=12ways$$$$$$$$Howisthiswrong?$$

Commented by Rasheed.Sindhi last updated on 13/May/18

$$\mathrm{After}\mathrm{choosing}\mathrm{a}\mathrm{particular}\mathrm{principal}$$$$\mathrm{and}\mathrm{a}\mathrm{particular}\mathrm{student},\mathrm{remaining}$$$$\mathrm{two}\mathrm{members}\mathrm{may}\mathrm{be}\left(\mathrm{i}\right)\mathrm{both}\mathrm{students}$$$$\left(\mathrm{ii}\right)\mathrm{both}\mathrm{principals}\mathrm{or}\left(\mathrm{iii}\right)\mathrm{a}\mathrm{principal}$$$$\mathrm{\&}\mathrm{a}\mathrm{students}.$$$$\mathrm{Do}\mathrm{you}\mathrm{think}\mathrm{that}\mathrm{the}\mathrm{remaining}\mathrm{two}$$$$\mathrm{members}\mathrm{must}\mathrm{contain}\mathrm{one}\mathrm{principal}$$$$\mathrm{and}\mathrm{one}\mathrm{student}?$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/May/18

$$principals\left(p\right)=4$$$$students\left(s\right)=5$$$$committeeoffour=4$$$$principals(p^{\ast },p_1,p_{2,}{p}_3)p^{\ast }=particularprincipal$$$$students(s^{\ast },s_1,s_{2,}{s}_3,s_4)s^{\ast }=particularstudent$$$$compositionof$$$$1)p^{\ast }{s}^{\ast }+twoprincipal=3_{C_2}$$$$2)p^{\ast }{s}^{\ast }+onestudent+oneprincipal=4_{C_1}\times 3_{C_1}$$$$3)p^{\ast }{s}^{\ast }+twostudents=4_{C_2}$$$$total=3_{C_2}+4_{C_1}\times 3_{C_1}+4_{C_2}$$$$=3+12+6=21$$

Commented by Rasheed.Sindhi last updated on 13/May/18

$$\mathrm{Your}\mathrm{answer}\mathrm{seems}\mathrm{correct}!$$

Commented by NECx last updated on 13/May/18

$$yeah...Ithinkthisisright.$$

Commented by NECx last updated on 13/May/18

$$Thanksforyourcontributions$$

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