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
310+3123×610×6+3×512×5=1860+1560Nowyoucanadd1860+1560=3360Simplify33÷360÷3=11201120
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)
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 !!! }