Menu Close

A-committee-of-4-is-to-be-formed-from-4-principals-and-5-students-In-how-many-ways-can-this-be-done-if-a-particular-student-and-a-principal-must-be-in-the-committee-




Question Number 34930 by NECx last updated on 13/May/18
A committee of 4 is to be formed  from 4 principals  and 5 students.In how many ways  can this be done if a particular  student and a principal must be  in the committee.
Acommitteeof4istobeformedfrom4principalsand5students.Inhowmanywayscanthisbedoneifaparticularstudentandaprincipalmustbeinthecommittee.
Commented by Rasheed.Sindhi last updated on 13/May/18
^• A particular principal  can be seleted  in  1   way.  ^• A particular student  can be seleted  in  1   way.  (Now there remain 4+5−2=7 persons)  ^• So third member may be selected in     7 ways.  (Now there remain 7−1=6 persons.  ^• The fourth member can be selected  in 6 ways  Total ways 1×1×7×6=42 ways.
Aparticularprincipalcanbeseletedin1way.Aparticularstudentcanbeseletedin1way.(Nowthereremain4+52=7persons)Sothirdmembermaybeselectedin7ways.(Nowthereremain71=6persons.Thefourthmembercanbeselectedin6waysTotalways1×1×7×6=42ways.
Commented by NECx last updated on 13/May/18
since a student and a principal must  be included then we have 3 principals  left to choose 1 from and 4 students  left to choose 1.  =C_1 ^3 ×C_1 ^4 =12 ways    How is this wrong?
sinceastudentandaprincipalmustbeincludedthenwehave3principalslefttochoose1fromand4studentslefttochoose1.=C31×C41=12waysHowisthiswrong?
Commented by Rasheed.Sindhi last updated on 13/May/18
After choosing a particular principal  and a particular student,remaining  two members may be (i) both students  (ii) both principals or ( iii)a principal  & a students.   Do you think that the remaining two  members must contain one principal  and one student?
Afterchoosingaparticularprincipalandaparticularstudent,remainingtwomembersmaybe(i)bothstudents(ii)bothprincipalsor(iii)aprincipal&astudents.Doyouthinkthattheremainingtwomembersmustcontainoneprincipalandonestudent?
Answered by tanmay.chaudhury50@gmail.com last updated on 13/May/18
principals(p)=4  students(s)=5  committee of four=4  principals(p^∗ ,p_1 ,p_(2,) p_3 )  p^∗ =particular principal  students(s^∗ ,s_1 ,s_(2 ,) s_3  ,s_4 ) s^∗ =particular student  composition of   1)p^∗ s^∗ +two principal=3_C_2    2)p^∗ s^∗ +one student+one principal=4_C_1  ×3_C_1    3)p^∗ s^∗ +two students=4_C_2    total=3_C_2  +4_C_1  ×3_C_1  +4_C_2    =3+12+6=21
principals(p)=4students(s)=5committeeoffour=4principals(p,p1,p2,p3)p=particularprincipalstudents(s,s1,s2,s3,s4)s=particularstudentcompositionof1)ps+twoprincipal=3C22)ps+onestudent+oneprincipal=4C1×3C13)ps+twostudents=4C2total=3C2+4C1×3C1+4C2=3+12+6=21
Commented by Rasheed.Sindhi last updated on 13/May/18
Your answer seems correct!
Youranswerseemscorrect!
Commented by NECx last updated on 13/May/18
yeah... I think this is right.
yeahIthinkthisisright.
Commented by NECx last updated on 13/May/18
Thanks for your contributions
Thanksforyourcontributions

Leave a Reply

Your email address will not be published. Required fields are marked *