Combinatrics

Stockmoose16 wrote:A box contains one dozen donuts. Four of the donuts are chocolate, four are glazed, and four are jelly. If two donuts are randomly selected from the box, one after the other, what is the probability that both will be jelly donuts?

1/11
1/9
1/3
2/3
8/9

I'm know to solve this problem using probability, you could just do 4/12*3/11 =1/11.

However, when I try to solve using combinatrics, I'm wondering if there's 2 different ways to do it:

VERSION #1
Number of total possibilities 12C2 = 66
Number of ways to pick 2 Jelly donuts: 4C2= 6

6/66=1/11

OR

VERSION #2
Number of total possibilities: 12C1 * 11C1= 12*11= 132
Number of ways to pick 2 Jelly donuts= 4C1 * 3C1= 4*3=12

12/132=1/11

Are both these combinatrics methods viable? If so, how come they both work, even though version #2 involves multiplication and version #1 does not?

canuckclint wrote:

Stockmoose16 wrote:A box contains one dozen donuts. Four of the donuts are chocolate, four are glazed, and four are jelly. If two donuts are randomly selected from the box, one after the other, what is the probability that both will be jelly donuts?

1/11
1/9
1/3
2/3
8/9

I'm know to solve this problem using probability, you could just do 4/12*3/11 =1/11.

However, when I try to solve using combinatrics, I'm wondering if there's 2 different ways to do it:

VERSION #1
Number of total possibilities 12C2 = 66
Number of ways to pick 2 Jelly donuts: 4C2= 6

6/66=1/11

OR

VERSION #2
Number of total possibilities: 12C1 * 11C1= 12*11= 132
Number of ways to pick 2 Jelly donuts= 4C1 * 3C1= 4*3=12

12/132=1/11

Are both these combinatrics methods viable? If so, how come they both work, even though version #2 involves multiplication and version #1 does not?

Yes you are absolutely right. The second way instead of having c 1.
Just have probability of picking 1st jelly donut * picking 2nd jelly donut
4/12 * 3/11 = 1/11, possibly version #2.5

Version 1 is not a good idea because (12 c 2) is very hard to compute by hand.