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 tried the above problem using Combinations, and I got the correct answer which is A.

Here's the method : 2* (6C1 * 1C1 * 5C2) + (6C2 * 2C1 * 4C1) = 240 {First part - Choose 1 card from the first suit, remaining two cards from the second suit; Second part - Choose 2 from each} {1C1 and 2c1 represent the same numbered card}


However, I crashed while using Permutations. Permutations - 12*1*10*8 = 960. (crash ) I don't know how I can arrive at 240 using this method. Really confused.

I tried a couple of other examples:

LEt's say there are only 4 cards of two suits each.

Combinations: 2* (4C1*1C1*3C1) + (4C2*2C1*2C1) = 48 {Same logic - choose 1 from the first suit; remainign three from the second suit; Second part - Choose 2 from each. }
Using permutations - 8* 1*6*4= 48*4...(crash )


I see that in both the examples, the "Combinations" number is "Permutations/4"...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.

Correct?

Bunuel,
Thanks for finding a similar question. However, I am not sure how to calculate the number of ways to obtain "a pair" using PErmutations. The question that you have posted is different - it asks us to calculate probability. This one is different because it asks us to calculate "combinations." I see that the author has just modified the same question. The source of this question is 700-800 series by MGMAT.....

Please help me I am really struggling...thanks

1) how many possible pairs? ==> 6

2) for each pair we need to find the no. of combinations of the other two cards. Since 2 cards are already chosen from the deck (the pair), there are 10 remaining cards.

C(10,2) = 45

3) 6*45 = 270 => this is no. of combinations of 4 cards that contain at least one pair. But this also includes the combinations with 2 pairs that should be eliminated.

4) how many combinations with 2 pairs? 6* C(5,1) = 30

5) 270-30=240.

Consider it like this-

We have to eliminate the case where two pairs are selected.

So C(6,1) is how u chose the first pair. Remaining number of pairs are 5.
So if u choose 2 cards that are of same number, then C(5,1) are the number of ways to do it.
Therefore C(6,1)*C(5,1)=30

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.



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.



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
_________________

Mike McGarry
Magoosh Test Prep

Mike,
Thanks for your elaborate reply. I believe that the analogy drawn by you is not 100% accurate. Mathematics and language are completely different. 2+2 = 4 everywhere in the Universe. Language is fluid. I see your point about Permutations and Combinations. The Combinations solution is computable. However, I wanted to "test" my permutations skills by trying the same problem with Permutations. I would really appreciate if you could help me with an alternate solution.

Here's what I could get :
Total number of permutations = 12*1*10*8=960 ways to choose a-pair. Now, these cards could be arranged in 4!/(2!*2!) = 6.

hence, total number of permutations = 960*6.
Now the correct answer = 960*6/24. I am not sure why we are supposed to divide by 4! = 24.

Help?