Mumbai

A job placement agency in Mumbai had to send ten students to five companies two to each. Two of the companies are in Mumbai and others are outside. Two of the students prefer to work in Mumbai while three prefer to work outside. In how many ways assignments can be made if preferences are to be satisfied?

Result

n =  682

Solution:

C2(7)=(72)=7!2!(7−2)!=7⋅62⋅1=21 C2(5)=(52)=5!2!(5−2)!=5⋅42⋅1=10  C2(2)=(22)=2!2!(2−2)!=11=1  a=(2+52)+(52)=(2+52)+10=31 C2(6)=(62)=6!2!(6−2)!=6⋅52⋅1=15  C2(4)=(42)=4!2!(4−2)!=4⋅32⋅1=6  b=(62)+(42)+(22)=15+6+1=22  n=a⋅ b=31⋅ 22=682C_{{ 2}}(7) = \dbinom{ 7}{ 2} = \dfrac{ 7! }{ 2!(7-2)!} = \dfrac{ 7 \cdot 6 } { 2 \cdot 1 } = 21 \ \\ C_{{ 2}}(5)=\dbinom{ 5}{ 2}=\dfrac{ 5! }{ 2!(5-2)!}=\dfrac{ 5 \cdot 4 } { 2 \cdot 1 }=10 \ \\ \ \\ C_{{ 2}}(2)=\dbinom{ 2}{ 2}=\dfrac{ 2! }{ 2!(2-2)!}=\dfrac{ 1 } { 1 }=1 \ \\ \ \\ a={ { 2+5 } \choose 2 } + { { 5 } \choose 2 }={ { 2+5 } \choose 2 } + 10=31 \ \\ C_{{ 2}}(6)=\dbinom{ 6}{ 2}=\dfrac{ 6! }{ 2!(6-2)!}=\dfrac{ 6 \cdot 5 } { 2 \cdot 1 }=15 \ \\ \ \\ C_{{ 2}}(4)=\dbinom{ 4}{ 2}=\dfrac{ 4! }{ 2!(4-2)!}=\dfrac{ 4 \cdot 3 } { 2 \cdot 1 }=6 \ \\ \ \\ b={ { 6 } \choose 2 } + { { 4 } \choose 2 } + { { 2 } \choose 2 }=15 + 6 + 1=22 \ \\ \ \\ n=a \cdot \ b=31 \cdot \ 22=682

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