Choosing balls probability

[DentalNucleicAcid]

In number 12 of QR in Destroyer we need to know the probability of choosing a red, a white, and a blue ball out of 3 red, 4 white, and 5 blue. After we obtain 1/22 to find the probability for one of the possible combinations of red, white and blue, why do we multiply by 3! instead of 12!/3! ? I understand that we're trying to obtain the rest of the combinations in addition to the 1/22 for one of them, but I would think that we'd do this by saying 12!/3! because we have 12 possible balls to choose from at first but then divide by 3! because we're making 3 selections, so we eliminate the repeated permutations.

 

[DAT Destroyer]

In number 12 of QR in Destroyer we need to know the probability of choosing a red, a white, and a blue ball out of 3 red, 4 white, and 5 blue. After we obtain 1/22 to find the probability for one of the possible combinations of red, white and blue, why do we multiply by 3! instead of 12!/3! ? I understand that we're trying to obtain the rest of the combinations in addition to the 1/22 for one of them, but I would think that we'd do this by saying 12!/3! because we have 12 possible balls to choose from at first but then divide by 3! because we're making 3 selections, so we eliminate the repeated permutations.

We multiply by 3!=6 because there are 6 different ways to pick 3 balls of different colors:
RWB, RBW, BRW, BWR, WBR, WRB.
Because the question did not specify the order we need to multiply by the number of possibilities to pick 3 different colors.

Hope this helps!

 

[DentalNucleicAcid]

We multiply by 3!=6 because there are 6 different ways to pick 3 balls of different colors:
RWB, RBW, BRW, BWR, WBR, WRB.
Because the question did not specify the order we need to multiply by the number of possibilities to pick 3 different colors.

Hope this helps!

Awesome, thank you.