combination problem--12 card deck.

[jsnipes]

Can someone show me how to solve this problem using factorials?

Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?
8/33
62/165
17/33
103/165
25/33

I seem to be struggling w combination problems largely in deciding what the best method of answering the question is and how to proceed (which is probably a product of my lack of understanding combination stuff in general). In retrospect I see that it makes sense to use the 'slot method' and to also find the probability of there NOT being any pairs. So basically:
(12/12)*(10/11)*(8/10)*(6/9)=16/33 and 1-16/33= 17/33 or C.

But I feel like I should be able to get to this same answer using some combination of x!/(x-n)!n! or something. Appreciate any help.

[shankar.ashwin]

Dont see a direct method of doing it, the way you did it seems the fastest way of doing it to me.

If you want to do it using combinations, it would be something like this.

Picking up 4 cards from 12 = 12C4 = 495

Picking up all 4 cards different = 2^4 (6C4) = 16*15 = 240

2^4 - Once you have all diff numbered cards, each of the card could be from either of the 2 decks, hence each numbered card could be from either Deck 1 or 2. Hence 6C4 is combination of picking up 4 diff numbered cards and each card should be inturn multiplied by 2.

P(Atleast 1 pair) = 1 - P(All diff) = 1 - (240/495) = 17/33

[tpr-becky]

Complex formulas that are difficult to apply can often get in the way of simpler solutions. It is better to have a basic way to approach problems.

In this case it is probability with at least - this means that you look for the formula:

1 - (probability of not getting what you want)

In this case it is 1 - (probability of getting no pairs).

the Probability of getting no pairs is multiple actions - so we will look at each action and then multiply.

first you can pick any card from the 12 - 12/12
Then you need a card that isn't a pair. subtract the other card that is a pair of this. 10/11
Then you need a card that isn't a pair of either: 8/10
Then you need a card that isn't a pair of any of the three: 6/9

Put together the formula: 1 - (12/12 * 10/11 * 8/10 * 6/9) = 17/33

All at least one problems will work this way.

[jsnipes]

thanks for the response becky. i think i try too hard to make everything fit into the n!/(n-k)!k! mold when i really just need to think about what the question is asking. that is certainly helpful to know that all 'at least one' problems can be solved using the slot method because that is very intuitive to me. i get hung up when i worry about how to parse the question into a simple n! etc formula

[fcabanski]

One of the traps of the GMAT and other similar tests is trying to make it seem that the "smart" student would know a simple formula. Another similar trap is that the "smart" students would be able to do the work in their heads.

There is nobody to impress. When you take the GMAT nobody will know if you counted on your fingers and toes. The best way to solve a problem is the quickest way, for you. That generally involves doing exactly what the OP did - consider the problem with simple steps.

[mbasubrosa]

Personally, I feel that your approach is fine for this question.
Yes, I see that you are trying to make such questions generic to be based on the combinations formula but it is much faster to use the mentioned technique for the given question