voodoochild wrote:
Randolph has a 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, there are 2 cards in the deck that have the same value. Randolph likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for a pair of cards that have the same value. How many such combinations are possible?
(A) 240
(B) 960
(C) 120
(D) 40
(E) 5760
I would use the FCP to solve this problem.
First two cards = the pair = there are six ways to get a pair.
Then two cards from the remaining 10 ---> 10C2 = 45
6*45 = 270
BUT, now we have to eliminate things that have been counted twice. What has been counted twice? The two pair combos. If I pick, say, two 4's as my first two cards, and then I happen to get two 2's as my last two cards, that will result in the same hand, the same combination, as if I pick the pair of 2's first and then the pair of 4s. How many pairs of pairs are there?
(6 choices for the first pair)*(5 for the remaining pair) = 30.
So, the 30 two-pair combinations have been counted twice, so subtract that number once.
270 - 30 = 240, the OA.
voodoochild wrote:
I tried the above problem using Combinations, and I got the correct answer which is A.
However, I crashed while using Permutations.
You know, from my perspective, this is analogous to someone saying ---
When I translated the French sentence with a French dictionary, it was no problem, but when I translated the French sentence with a German dictionary, I had all kinds of problems!
Combinations and permutations are two very different things. If a problem, such as this one, is about combinations --- that is to say, only the final grouping matters, and order does not matter in the least --- then a combinations approach is correct and permutations approach is not correct, unless you do something to the permutations, like nPr/r!, to make it a combinations approach in disguise.
Here's my advice. If the problem is about combinations, don't use permutations. Period.
voodoochild wrote:
Is there a rule to quickly relate Permutations with Combinations? I know that nCr = nPr/r!; However, I believe that this theorem is not applicable here.
nCr = nPr/r! is always true, absolutely true, 100% of the time. I believe you calculated the permutations incorrectly --- to be honest, I have no idea what you are doing in your permutations calculations: I do not recognize them either as permutations or as anything related to this problem. The underlying problem I believe is, again, this problem is by its very nature of combination problem. It's not at all clear to me how permutations would be the least bit helpful in solving it. It's like trying to use German to translate a French sentence.
Does all this make sense?
Mike