Answer to Question #283541 in Discrete Mathematics for Vamshi

Question #283541

A committee of 8 is to be formed from 16 men and 10 women.


In how many ways can the committee be formed if


i) there are no restrictions.


ii) there must be 4 men and 4 women.


iii) there should be an even number of women.


iv) there are more women than men.


v) there are atleast 6 men.

1
Expert's answer
2021-12-30T10:43:09-0500

i) there are no restrictions:

nCr=n!r!(nr)!n\raisebox{0.25em}{$C$}r=\frac{n!}{r!(n-r)!}

26C8=26!8!(268)!=26!8!×18!=156227526\raisebox{0.25em}{$C$}8=\frac{26!}{8!(26-8)!}=\frac{26!}{8!\times18!}=1562275


ii) there must be 4 men and 4 women:

(16C4)(10C4)=(16!4!(164)!)(10!4!(104)!)(16\raisebox{0.25em}{$C$}4)(10\raisebox{0.25em}{$C$}4)=(\frac{16!}{4!(16-4)!})(\frac{10!}{4!(10-4)!})

(16C4)(10C4)=(16!4!×12!)(10!4!×6!)=382200(16\raisebox{0.25em}{$C$}4)(10\raisebox{0.25em}{$C$}4)=(\frac{16!}{4!\times12!})(\frac{10!}{4!\times6!})=382200


iii) there should be an even number of women:

=(16C6)(10C2)+(16C4)(10C4)+(16C2)(10C6)+(16C0)(10C8)=(16\raisebox{0.25em}{$C$}6)(10\raisebox{0.25em}{$C$}2)+(16\raisebox{0.25em}{$C$}4)(10\raisebox{0.25em}{$C$}4)+(16\raisebox{0.25em}{$C$}2)(10\raisebox{0.25em}{$C$}6)+(16\raisebox{0.25em}{$C$}0)(10\raisebox{0.25em}{$C$}8)

=(16!6!(166)!)(10!4!(104)!)+(16!4!(164)!)(10!4!(104)!)+(16!2!(162)!)(10!6!(106)!)+(16!0!(160)!)(10!8!(108)!)=(\frac{16!}{6!(16-6)!})(\frac{10!}{4!(10-4)!})+(\frac{16!}{4!(16-4)!})(\frac{10!}{4!(10-4)!})+(\frac{16!}{2!(16-2)!})(\frac{10!}{6!(10-6)!})+(\frac{16!}{0!(16-0)!})(\frac{10!}{8!(10-8)!})

=(16!6!(10)!)(10!4!(6)!)+(16!4!(12)!)(10!4!(6)!)+(16!2!(14)!)(10!6!(4)!)+(16!0!(16)!)(10!8!(2)!)=(\frac{16!}{6!(10)!})(\frac{10!}{4!(6)!})+(\frac{16!}{4!(12)!})(\frac{10!}{4!(6)!})+(\frac{16!}{2!(14)!})(\frac{10!}{6!(4)!})+(\frac{16!}{0!(16)!})(\frac{10!}{8!(2)!})

=360360+382200+25200+45=767805=360360+382200+25200+45=767805


iv) there are more women than men:

=(16C0)(10C8)+(16C1)(10C7)+(16C2)(10C6)+(16C3)(10C5)=(16\raisebox{0.25em}{$C$}0)(10\raisebox{0.25em}{$C$}8)+(16\raisebox{0.25em}{$C$}1)(10\raisebox{0.25em}{$C$}7)+(16\raisebox{0.25em}{$C$}2)(10\raisebox{0.25em}{$C$}6)+(16\raisebox{0.25em}{$C$}3)(10\raisebox{0.25em}{$C$}5)

=(16!0!(160)!)(10!8!(108)!)+(16!1!(161)!)(10!7!(107)!)+(16!2!(162)!)(10!6!(106)!)+(16!3!(163)!)(10!5!(105)!)=(\frac{16!}{0!(16-0)!})(\frac{10!}{8!(10-8)!})+(\frac{16!}{1!(16-1)!})(\frac{10!}{7!(10-7)!})+(\frac{16!}{2!(16-2)!})(\frac{10!}{6!(10-6)!})+(\frac{16!}{3!(16-3)!})(\frac{10!}{5!(10-5)!})

=(16!!0(16)!)(10!8!(2)!)+(16!1!(15)!)(10!7!(1)!)+(16!2!(14)!)(10!6!(4)!)+(16!3!(13)!)(10!5!(5)!)=(\frac{16!}{!0(16)!})(\frac{10!}{8!(2)!})+(\frac{16!}{1!(15)!})(\frac{10!}{7!(1)!})+(\frac{16!}{2!(14)!})(\frac{10!}{6!(4)!})+(\frac{16!}{3!(13)!})(\frac{10!}{5!(5)!})

=45+1920+25200+141120=168285=45+1920+25200+141120=168285


v) there are atleast 6 men:

=(16C6)(10C2)+(16C7)(10C1)+(16C8)(10C0)=(16\raisebox{0.25em}{$C$}6)(10\raisebox{0.25em}{$C$}2)+(16\raisebox{0.25em}{$C$}7)(10\raisebox{0.25em}{$C$}1)+(16\raisebox{0.25em}{$C$}8)(10\raisebox{0.25em}{$C$}0)

=(16!6!(166)!)(10!2!(102)!)+(16!7!(167)!)(10!1!(101)!)+(16!8!(168)!)(10!0!(100)!)=(\frac{16!}{6!(16-6)!})(\frac{10!}{2!(10-2)!})+(\frac{16!}{7!(16-7)!})(\frac{10!}{1!(10-1)!})+(\frac{16!}{8!(16-8)!})(\frac{10!}{0!(10-0)!})

=(16!6!(10)!)(10!2!(8)!)+(16!7!(9)!)(10!1!(9)!)+(16!8!(8)!)(10!0!(10)!)=(\frac{16!}{6!(10)!})(\frac{10!}{2!(8)!})+(\frac{16!}{7!(9)!})(\frac{10!}{1!(9)!})+(\frac{16!}{8!(8)!})(\frac{10!}{0!(10)!})

=360360+114400+12870=487630=360360+114400+12870=487630



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