In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls ?

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Solution

There are 5 boys and 4 girls

The team consists of 3 boys and 3 girls.

Number of ways to from the leom

$=^{5} C_{3} \times^{4} C_{3}$

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

$= \frac{5 \times 4}{2} \times 4$

$= 40$

Number of ways = 40

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In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls ?

There are 5 boys and 4 girls The team consists of 3 boys and 3 girls. Number of ways to from the leom =5C3×4C3 =5!3!2!×4!3! =5×42×4 =40 Number of ways = 40

There are 5 boys and 4 girls

The team consists of 3 boys and 3 girls.

Number of ways to from the leom

$=^{5} C_{3} \times^{4} C_{3}$

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

$= \frac{5 \times 4}{2} \times 4$

$= 40$

Number of ways = 40