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

Question

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Answer 1

Total number of girls in the group $= 4$

Total number of boys in the group $= 7$

Probability of selecting atleast one boy and one girl =?

By considering following ways we can select

i) $1$ girls $+ 4$ boys $\Rightarrow^{4} C_{1} \times^{7} C_{4}$

ii) $2$ girls $+ 3$ boys $\Rightarrow^{4} C_{2} \times^{7} C_{3}$

iii) $3$ girls $+ 2$ boys $\Rightarrow^{4} C_{3} \times^{7} C_{2}$

iv) $4$ girls $+ 1$ boy $\Rightarrow^{4} C_{4} \times^{7} C_{1}$

Total number of ways $= (^{4} C_{1} \times^{7} C_{4}) + (^{4} C_{2} \times^{7} C_{3}) + (^{4} C_{3} \times^{7} C_{2}) + (^{4} C_{4} \times^{7} C_{1})$

$= 140 + 210 + 84 + 7$

$= 441$

Answer 2

Total number of girls in the group

= 4

Total number of boys in the group

= 7

Probability of selecting atleast one boy and one girl =?

By considering following ways we can select

i)

1

girls

+ 4

boys

\Rightarrow^{4} C_{1} \times^{7} C_{4}

ii)

2

girls

+ 3

boys

\Rightarrow^{4} C_{2} \times^{7} C_{3}

iii)

3

girls

+ 2

boys

\Rightarrow^{4} C_{3} \times^{7} C_{2}

iv)

4

girls

+ 1

boy

\Rightarrow^{4} C_{4} \times^{7} C_{1}

Total number of ways

= (^{4} C_{1} \times^{7} C_{4}) + (^{4} C_{2} \times^{7} C_{3}) + (^{4} C_{3} \times^{7} C_{2}) + (^{4} C_{4} \times^{7} C_{1})

= 140 + 210 + 84 + 7

= 441