Randolph has a deck of 12 playing cards made up of only 2

Question

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?

mikemcgarry · Accepted Answer

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 :-)

Bunuel · Answer

Basically the same question as this one: bill-has-a-small-deck-of-12-playing-cards-made-up-of-only-2-suits-of-6-cards-each-96078.html 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? A. 8/33 A. 62/165 C. 17/33 D. 103/165 E. 25/33 Let's calculate the opposite probability ans subtract this value from 1. Opposite probability would be that there will be no pair in 4 cards, meaning that all 4 cards will be different: \frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33} . C^4_6 - # of ways to choose 4 different cards out of 6 different values; 2^4 - as each of 4 cards chosen can be of 2 different suits; C^4_{12} - total # of ways to choose 4 cards out of 12. So P=1-\frac{16}{33}=\frac{17}{33} . Or another way: We can choose any card for the first one - \frac{12}{12} ; Next card can be any card but 1 of the value we'v already chosen - \frac{10}{11} (if we've picked 3, then there are one more 3 left and we can choose any but this one card out of 11 cards left); Next card can be any card but 2 of the values we'v already chosen - \frac{8}{10} (if we've picked 3 and 5, then there are one 3 and one 5 left and we can choose any but these 2 cards out of 10 cards left); Last card can be any card but 3 of the value we'v already chosen - \frac{6}{9} ; P=\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=\frac{16}{33} . So P=1-\frac{16}{33}=\frac{17}{33} - the same answer as above. Answer: C.

EvaJager · Answer

If "How many such combinations are possible?" refers to 4 cards containing at least one pair of cards with the same value, then the correct answer is not listed.

The total number of possibilities to chose any 4 cards, regardless the order in which they were chosen, is given by  12C4=\frac{12*11*10*9}{4!}=\frac{990}{2}=495 .
The number of possibilities to chose a set of 4 cards containing no pair of equal numbers, regardless the order in which they were chosen, is given by  12*10*8*6/4!=240 . First card 12 possibilities, second any card with a number different from the first one, so  12-2 =10  possibilities, for the third one  12-4=8  possibilities, and for the fourth card  12-6=6  possibilities. Divide by  4! , since order doesn't matter.

So, the number of possibilities to get at least one pair of cards with equal numbers among the chosen 4 is  495 - 240 = 255 , which is not listed as an answer!

In its present form, this cannot be a real GMAT test question, the wording is really poor.

EvaJager · Answer

In my previous post I interpreted the question as the number of possibilities to get at least one pair, which in fact means either one or two pairs.
I got 255 possibilities.

The number of possibilities to get two pairs is 6C2=6*5/2=15, we just have to choose two pairs out of 6.
Obviously, the number of possibilities for exactly one pair is 255 - 15 = 240.

Here, I am presenting an approach to obtain directly this number:

For the one pair, we have 6 possibilities to choose from. 
For the other two cards,  not to form a pair : we have for the first card 10 possibilities, then for the second 8 possibilities (one pair already chosen, and the card cannot be the pair of the previous one).
This would give 10*8=80 possibilities, but we have to divide by 2, as order doesn't matter, and we are left we 80/2 = 40.

Thus, the final number of possibilities is 6*40 = 240.

Answer A

voodoochild · Answer

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?

EvaJager · Answer

Because when you have chosen your cards, you took into account order: first card, second card,...
So, for a given set of 4 cards ABCD, you counted all possible permutations ACBD, ADBC, ...
That's why you have to divide by 4! which is exactly the number of different arrangements of the 4 cards.

deya · Answer

The above explanations are not clear to me. I have tried hard to make myself understand but I am failed in doing that. Kindly make me understand in another possible way. Thank You in advance :-D

mikemcgarry · Answer

Dear deya , I'm happy to respond. :-)

My friend, with all due respect, this is not a very good question. Make you understand? I understand the problem, but how can I "make you understand" if I know nothing about you? Please explain, in as much detail as possible: (1) what you do understand about this problem? (2) what steps or statements in the above solutions do you not understand? (3) what did you think of and what did you try when you approached the problem? Part of being a successful GMAT student is learning to ask excellent questions. It takes a great deal more work, but if you put in the effort to ask excellent questions, it will actually prime your mind to understand more deeply. Learning is not just about getting the information: much more importantly, learning is about preparing you mind to absorb the information. Does all this make sense? Mike :-)

usre123 · Answer

Could someone please help with this? thanks

KarishmaB · Answer

Assuming the question means "only one pair" (not a valid assumption though)

You have ordered the cards. Note that you need to select two kinds of cards - one a pair and another a set of two cards which are different.
You can select the pair in 6 ways (pair of 1 or pair of 2 or pair of 3 etc). Another way of saying this is: select the first card of the pair in 12 ways and then the second card in 1 way only. But since there is no first-second arrangement, divide 12 by 2! to get 6.
You select the dissimilar cards in 10*8 ways - 10 ways for the first one and 8 ways for the second one. Again, there is no first-second so you divide this by 2 to get 10*8/2 = 40 ways.

Total you get 6*40 = 240 ways

ScottTargetTestPrep · Answer

Solution:

Since there are 12 cards, the number of ways one can choose 4 cards is:

12C4 = (12 x 11 x 10 x 9)/(4 x 3 x 2) = 11 x 5 x 9 = 495

When 4 cards are chosen, in terms of the number of pairs of cards that have the same value, there could be 0, 1, or 2 pairs. If we can determine the number of ways the 4 cards have 0 pairs and 2 pairs, then we can subtract those two results from 495 to obtain the number of ways the 4 cards would have (exactly) 1 pair.

The number of ways 4 cards have no pair is:

(12 x 10 x 8 x 6) / (4 x 3 x 2) = 5 x 8 x 6 = 240

(Note: in the above calculation, 12 is the number of ways one can choose the first card, 10 the second card, 8 the third card, and 6 the fourth card. However, since the order of the 4 cards doesn’t matter, we need to divide by 4! or 4 x 3 x 2.)

The number of ways 4 cards have 2 pairs is:

6C2 = (6 x 5)/2 = 15

(Note: Since we are really choosing 2 pairs from the available 6 pairs where order doesn’t matter.)

Therefore, the number of ways 4 cards have exactly 1 pair is 495 - 240 - 15 = 240.

Answer: A

mihir51modi · Answer

Total possible combination is 12C4= 495
Total possible combination when cards are not same = (12*10*8*6)/4!= 240
So 2 cards are same combination is 495-240= 255.
Answer is 255 which is not in options.

Posted from my mobile device

Adarsh_24 · Answer

Question maker meant only 1 pair allowed I think. But, the wording is just plain wrong.

bumpbot · Answer

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Randolph has a deck of 12 playing cards made up of only 2

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