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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.

$${A}\:{committee}\:{of}\:\mathrm{4}\:{is}\:{to}\:{be}\:{formed} \\ $$$${from}\:\mathrm{4}\:{principals} \\ $$$${and}\:\mathrm{5}\:{students}.{In}\:{how}\:{many}\:{ways} \\ $$$${can}\:{this}\:{be}\:{done}\:{if}\:{a}\:{particular} \\ $$$${student}\:{and}\:{a}\:{principal}\:{must}\:{be} \\ $$$${in}\:{the}\:{committee}. \\ $$$$ \\ $$

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.

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

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?

$${since}\:{a}\:{student}\:{and}\:{a}\:{principal}\:{must} \\ $$$${be}\:{included}\:{then}\:{we}\:{have}\:\mathrm{3}\:{principals} \\ $$$${left}\:{to}\:{choose}\:\mathrm{1}\:{from}\:{and}\:\mathrm{4}\:{students} \\ $$$${left}\:{to}\:{choose}\:\mathrm{1}. \\ $$$$=\overset{\mathrm{3}} {{C}}_{\mathrm{1}} ×\overset{\mathrm{4}} {{C}}_{\mathrm{1}} =\mathrm{12}\:{ways} \\ $$$$ \\ $$$${How}\:{is}\:{this}\:{wrong}? \\ $$

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?

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

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

principals(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

$${principals}\left({p}\right)=\mathrm{4} \\ $$$${students}\left({s}\right)=\mathrm{5} \\ $$$${committee}\:{of}\:{four}=\mathrm{4} \\ $$$${principals}\left({p}^{\ast} ,{p}_{\mathrm{1}} ,{p}_{\mathrm{2},} {p}_{\mathrm{3}} \right)\:\:{p}^{\ast} ={particular}\:{principal} \\ $$$${students}\left({s}^{\ast} ,{s}_{\mathrm{1}} ,{s}_{\mathrm{2}\:,} {s}_{\mathrm{3}} \:,{s}_{\mathrm{4}} \right)\:{s}^{\ast} ={particular}\:{student} \\ $$$${composition}\:{of}\: \\ $$$$\left.\mathrm{1}\right){p}^{\ast} {s}^{\ast} +{two}\:{principal}=\mathrm{3}_{{C}_{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right){p}^{\ast} {s}^{\ast} +{one}\:{student}+{one}\:{principal}=\mathrm{4}_{{C}_{\mathrm{1}} } ×\mathrm{3}_{{C}_{\mathrm{1}} } \\ $$$$\left.\mathrm{3}\right){p}^{\ast} {s}^{\ast} +{two}\:{students}=\mathrm{4}_{{C}_{\mathrm{2}} } \\ $$$${total}=\mathrm{3}_{{C}_{\mathrm{2}} } +\mathrm{4}_{{C}_{\mathrm{1}} } ×\mathrm{3}_{{C}_{\mathrm{1}} } +\mathrm{4}_{{C}_{\mathrm{2}} } \\ $$$$=\mathrm{3}+\mathrm{12}+\mathrm{6}=\mathrm{21} \\ $$

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

Your answer seems correct!

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

Commented by NECx last updated on 13/May/18

yeah... I think this is right.

$${yeah}...\:{I}\:{think}\:{this}\:{is}\:{right}. \\ $$

Commented by NECx last updated on 13/May/18

Thanks for your contributions

$${Thanks}\:{for}\:{your}\:{contributions} \\ $$

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