a class of juniors has 12 girls and 10 boys. In how many ways can a committee of 3 girls and 3 boys be chosen?

We have 12 girls and we can pick any three.....and we have 10 boys and we can pick any three

So  C(12,3) * C(10,3)  = 26,400 ways

Headingnorth's answer would have been correct if he/she had divided each of the products in both sets of parentheses by 3!........The reason for this is, for example, the set of {Ted, Bob, Dave} would be the same as (Bob, Ted, Dave}......and there are 3! = 6 arrangements of this set....but....it's basically the same set.......without the division by 3! in both cases, we're "overcounting" by a factor of 6 x 6  = 36...... 

{That's OK, Headingnorth, it was a really good try  !!!  .....I still gave you 3 points  !!! }

Letme take a guess.

For girls. 3/12

For boys. 3/10

3/10+3/12

LCD

60

$$\\\frac{3}{10}+\frac{3}{12}\\\\
\frac{3\times6}{10\times6}+\frac{3\times5}{12\times5}\\\\
=\frac{18}{60}+\frac{15}{60}\\\\
Now\;you\;can\;add\\\\
\frac{18}{60}+\frac{15}{60}=\frac{33}{60}\\\\
Simplify\frac{33\div\;3}{60\div\;3}=\frac{11}{20}\\\\
\frac{11}{20}$$

Hey!!!

I´m not totally sure about this but I think it make sense :/

We can pick 3 girls out of 12 in 12*11*10 ways.

(first we can choose from 12 girls, when we picked one we got 11 to choose from and so on...)

In the same way of thinking we can pick 3 boys in 10*9*8 different ways.

then we multiplie the numbers of different ways to combine the girls with the number of different ways to combine the boys.

so we got (12*11*10)*(10*9*8)

sounds more reasonable:)

thanks:)

Thanks CPhill and HeadingNorth,

MathsGod1, you answer had nothing to do with the question LOL 

BUT

You did the fraction addition perfectly  

3 points for all of you from me