Answer to Question #283541 in Discrete Mathematics for Vamshi

Question

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.

Answer 1

i) there are no restrictions:

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

$26\raisebox{0.25em}{$C$}8=\frac{26!}{8!(26-8)!}=\frac{26!}{8!\times18!}=1562275$

ii) there must be 4 men and 4 women:

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

$(16\raisebox{0.25em}{$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:

$=(16\raisebox{0.25em}{$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$}8)$

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

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

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

iv) there are more women than men:

$=(16\raisebox{0.25em}{$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$}5)$

$=(\frac{16!}{0!(16-0)!})(\frac{10!}{8!(10-8)!})+(\frac{16!}{1!(16-1)!})(\frac{10!}{7!(10-7)!})+(\frac{16!}{2!(16-2)!})(\frac{10!}{6!(10-6)!})+(\frac{16!}{3!(16-3)!})(\frac{10!}{5!(10-5)!})$

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

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

v) there are atleast 6 men:

$=(16\raisebox{0.25em}{$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$}0)$

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

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

$=360360+114400+12870=487630$