Notice that the question doesn't say you must choose R first, then G, then B. The reason for this is that if you use the probabilities approach, you need to take into account all the ways in which a favourable outcome can occur.

You are correct that you need to multiply by 6. My question is why do I need to consider these possibilities? in either case I have 1 R 1 G and 1 B. which in all are 6 possibilitie so 6*1/21 = 2/7. I believe this is because I am not considering the other possibilities like rbg, grb,gbr, brg,bgr etc. Why do this ans not ,match the previous one? However, there is also one more approach. Now one way to solve this is by the following method:. What is the probability that all 3 cards are a different color? 3 cards are randomly drawn without replacement from the deck. I need yur help to resolve one basic question that i face in probability.Ī deck of 9 cards contains 2 red cards, 3 blue cards, and 4 green cards. Also, if you can 'map' the problem to one of the two scenarios above, you should have no problem picking the correct formula to use.Hi Ursula, Once you understand the concept, you can reconstruct the formulas very quickly, even if you forget them under pressure. I completely agree with you that understanding the concept behind it is a much more solid approach than just memorizing the formulas. Also, if you can 'map' the problem to one of the two scenarios above, you should have no problem picking the correct formula to use. That's why the combination formula includes the division by n!, soĬ(5,3) = P(5,3) / 3! = (5 x 4 x 3) / (3 x 2) = 10 These selections all result in the same committee, so they are all equivalent to a single combination. This is simply the factorial of the number of committee members, in this case 3!įor example, consider these 6 permutations:

We need to divide by the number of possible ways in which a particular committee could have been picked. So the straightforward selection process we used to award prizes for the race needs to be adapted a bit. For example, it doesn't matter whether your selection is person A, then C, then D or whether it is C, then D, then A. You could do your selection in the same way as above, but you would find that some of these permutations give you the same committee. Now think of the same 5 people, but your task this time is to form a committee of 3 (with no special roles, just equal members). In this case, instead of the complete factorial of 5 x 4 x 3 x 2 x 1, you only have three prizes to be awarded, so you stop after the third term. That's why I prefer to think of it as simply an incomplete factorial where you multiply out only as many terms as you have selections. Notice that the effect of the denominator in the formula is just to cancel out the last two terms from the numerator. You can get this result using the permutation formula: Let's say you have 5 people who are running a race, and you want to know in how many ways three prizes (gold, silver, and bronze medal) could be awarded.Īny of the 5 could win the gold, any of the remaining 4 could win silver, and any of the other 3 could win bronze. I came up with this mnemonic device to keep these straight: Therefore we divide the number by the number of repetion that will take plae.įirst of all, you got the Combination and Permutation mixed up - but I guess that was the whole point of your post ) Had there been 4 different balls tha answer would have been 4!=24.Īnd hence there is repetioin of some sequence.

Now you have to place them in 4 positions. Now you have to place them in 3 different holes.īut with 3 same balls you would get only 1 combination.Ĭombination takes care by dividing the permutation with the number or repetitions that occur.

Suppose you have 3 balls blue green and red. In case of combination things that are to be distributed are not the same. But you can definately place them in 6 ways Whatever you do they would look like the same. Suppose you have 3 red balls and you would like to place them in three holes. permutation is done when things are same.

Permutation vs combination formula