Question

A committee of $10$ is to be formed from $8$ teachers and $12$ students of whom $4$ are girls. In how many ways this can be done, so that the committee contains atleast four of either groups (teachers and students) and atleast $2$ girls and atleast $2$ boys are in the committee.

Answer 1

$8teachersand12students(4girls+8boys)$

$Required:-committeeof10withatleast4of$

$eithergroupsandatleast2girls$

$and2boys.$

$case\left(i\right):4students,6teachers$

$numberofwaysofselecting6teachers$

$outof8{=}_6^{8}C$

$4studentsmustbesuchthat2aregirlsand$

$2boys.So,numberofways{=}_2^{4}C{.}_2^{8}C$

$numberofwaysofselecting{=}_6^{8}C{.}_2^{4}C{.}_2^{8}C$

$case\left(ii\right):5students,5teachers$

$Numberofwaystoselect5teachers{=}_5^{8}C$

$5studentscanbeeither3girls,2boys$

$so,Numberofways{=}_5^{8}C{(}_2^{4}C{.}_2^{8}C{+}_2^4{C.}_3^{8}C)$

$case\left(iii\right):6students,4teachers$

$6studentscaneitherbe2girls,4boysor3girls,3boys$

$or4girls,2boys$

$Numberofwaystoselect4teachers{=}_4^{8}C$

$So,numberofways{=}_4^{8}C{(}_2^{4}C{.}_4^{8}C{+}_3^{4}C{.}_3^{8}C{+}_4^4{C.}_2^{8}C)$

$totalnumberofways{=}_6^{8}C{.}_2^4{C.}_2^8{C+}_5^8{C(}_3^{4}C{.}_2^8{C+}_2^{4}C{.}_3^8{C)+}_4^{8}C{(}_2^{4}C{.}_4^8{C+}_3^4{C.}_3^{8}C{+}_4^{4}C{.}_2^{8}C)$

$=4704+56(4\times 28+6\times 56)+70(6\times 70+4\times 56+1\times 28)$

$=76,832ways$