Question

$1)$ A group consists of $4$ girls and $7$ boys. In how many ways can a team of $5$ members be selected if the team has
$i)$ No girl$?$

$2)$ A group consists of $4$ girls and $7$ boys. In how many ways can a team of $5$ members be selected if the team has
$ii)$ At least one boy and one girl$?$

$3)$ A group consists of $4$ girls and $7$ boys. In how many ways can a team of $5$ members be selected if the team has
$iii)$ At least $3$ girls$?$

Answer 1

$1)$ Finding number of ways to select a team with no girl.
Number of Girls $=4$

Number of Boys $=7$

Number of ways to choose boys $={}^7{C}_5$

Number of ways to choose girl $={}^4{C}_0$

$\therefore$ Total number of ways $={}^4{C}_0\times {}^7{C}_5$

$=\frac{4!}{0!4!}\times \frac{7!}{5!(7-5)!}$

$=\frac{7\times 6\times 5!}{5!\times 2!}$

$=\frac{7\times 6}{2\times 1}$

$=21$

$2)$ Step $1$
Team has $1$ boy and $4$ girls

Number of ways of selecting $1$ boy $={}^7{C}_1$

Number of ways of selecting $4$ girls $={}^4{C}_4$

Number of ways of selecting $1$ boy and $4$ girls $={}^7{C}_1\times {}^4{C}_4$

$=\frac{7!}{1!(7-1)!}\times \frac{4!}{4!(4-4)!}$

$=\frac{7\times 6!}{6!}\times \frac{4!}{4!0!}$

$=7\times 1$

$=7$

Step $2$

Team has $2$ boys and $3$ girls

Number of ways of selecting $2$ boys $={}^7{C}_2$

Number of ways of selecting $3$ girls $={}^4{C}_3$

Number of ways of selecting $2$ boys and $3$ girls $={}^7{C}_2\times {}^4{C}_3$

$=\frac{7\times 6\times 5!}{2!\times (7-2)!}\times \frac{4\times 3!}{3!(4-3)!}$

$=\frac{7\times 6\times 5!}{2\times 1\times 5!}\times \frac{4}{1!}$

$=21\times 4$

$=84$ ways

Step $3$

Team has $3$ boys and $2$ girls

Number of ways of selecting $3$ boys $={}^7{C}_3$

Number of ways of selecting $2$ girls $={}^4{C}_2$

$\therefore$ Number of ways of selecting $3$ boys and $2$ girls $={}^7{C}_3\times {}^4{C}_2$

$=\frac{7\times 6\times 5\times 4!}{(7-3)!3!}\times \frac{4\times 3\times 2!}{2!(4-2)!}$

$=\frac{7\times 6\times 5\times 4!}{4!\times 3\times 2\times 1}\times \frac{4\times 3\times 2}{2\times 2}$

$=35\times 6$

$=210$ ways

Step $4$

Team has $4$ boys and $1$ girl

Number of ways of selecting $4$ boys $={}^7{C}_4$

Number of ways of selecting $1$ girl $={}^4{C}_1$

$\therefore$ Number of ways of selecting $4$ boys and $1$ girl $={}^7{C}_4\times {}^4{C}_1$

$=\frac{7\times 6\times 5\times 4!}{4!3!}\times \frac{4!}{3!}=\frac{7\times 6\times 5}{3!}\times \frac{4\times 3!}{3!}$

$=7\times 5\times 4$

$=140$

Step $5$
Finding total number of ways.

So,
Total Number of ways $=7+84+210+140$

$=441$

$3)$ Step $1$
Team has $3$ girls and $2$ boys.

Number of ways to select $3$ girls $={}^4{C}_3$

Number of ways to select $2$ boys$={}^7{C}_2$

$\therefore$ Number of ways of select $3$ girls and $2$ boys

$={}^4{C}_3\times {}^7{C}_2=\frac{4!}{(4-3)!3!}\times \frac{7!}{(7-2)!2!}$

$=\frac{4\times 3!}{1!\times 3!}\times \frac{7\times 6\times 5!}{5!\times 2!}$

$=4\times \frac{7\times 6}{2\times 1}$

$=3\times 4\times 7$

$=84$ ways.

Step $2$
Team has $4$ girls and $1$ boy.

Number of ways to select $4$ girls $={}^4{C}_4$

Number of ways to select $1$ boy$={}^7{C}_1$

$\therefore$ Number of ways of select $4$ girls and $1$ boy

$={}^4{C}_4\times {}^7{C}_1$

$=\frac{4!}{4!(4-4)!}\times \frac{7!}{(7-1)!1!}$

$=\frac{1}{0!}\times \frac{7\times 6!}{6!\times 1!}$

$=7$

Step $3$

Finding total number of ways.

Total number of ways $=84+7$

$=91$