Commitee

A class consists of 6 males and 7 females. How many committees of 7 are possible if the committee must consist of 2 males and 5 females?

Result

n =  315

Solution:

C2(6)=(62)=6!2!(62)!=6521=15 C5(7)=(75)=7!5!(75)!=7621=21  n=C2(6)C5(7)=315C_{{ 2}}(6) = \dbinom{ 6}{ 2} = \dfrac{ 6! }{ 2!(6-2)!} = \dfrac{ 6 \cdot 5 } { 2 \cdot 1 } = 15 \ \\ C_{{ 5}}(7) = \dbinom{ 7}{ 5} = \dfrac{ 7! }{ 5!(7-5)!} = \dfrac{ 7 \cdot 6 } { 2 \cdot 1 } = 21 \ \\ \ \\ n = C_{ 2}(6) \cdot C_{ 5}(7) = 315



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