russ9 wrote:
The example you had about 5 seats and 3 people, that required us to "arrange" in a sense but a very similar problem where we had to choose 3 out of 5, we had to "unarrange". I guess my biggest disconnect is what was the trigger that we had to arrange? Was it because they had given us seats?
After selection, if you are going to assign them a place/number/position, you need permutation. If you are just going to carry on with a bunch, this is combination.
5 friends and 3 seats example needed you to select 5 friends and then make them sit on 3 distinct seats - permutation
When you had to select 3 friends out of 5 to take them to a vacation, you just need a bunch. You are not allotting them any positions - combination
russ9 wrote:
Is it safe to say that when the GMAT wants us to PUT something/something IN/ON something -- that usually means that it's an arrange problem, hence a permutation problem?
A random question that I just thought might be relevant? I'm going to paraphrase here but it went something like, "There are 3 red marbles and 7 blue. We pick 3, what is the probability of picking at least 2 blue". I understand that we are adding a probability scenario here but if we just talk about the "favorable outcomes/numerator" and talk about the different combinations, why is it that we have to "arrange" BBR, BRB, RBB when we use the probability equation -- (7/10)(6/9)(3/8) + (7/10)(3/9)(2/8) + ... BUT if we use a Combination/Permutation equation, we don't have to arrange, meaning, we can solve by doing (7C2)(3c1) but we don't need to account for if it's BBR, BRB etc.
Sorry for the long winded question but I hope this helps clarify the topic for other users as well.
Thanks a ton,
Russ
Very hard to generalize the "IN/ON" statement but yeah, I would guess that would be true in most cases.
You consider all arrangements such as BBR/BRB etc because these are the numerous ways in which you can pick marbles.
I am assuming that you meant to write "the probability of picking 2 Blue and 1 Red marbles".
In the combination approach, you ignore arrangements because it is a probability question. You CAN arrange in both favorable cases as well as total cases. The ratio of Favorable/Total will stay the same. To arrange the 3 balls you will multiply the numerator by 3!. To arrange the 3 balls in total cases, you will multiply the denominator by 3! too. The 3!s get canceled. That is why you can ignore the arrangements.
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