Tracy 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, there are 2 cards in the deck that have the same 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 a pair of cards that have the same value?
a)1/33
b)16/33
c)17/33
d)32/33
e)1
OA = B
Solution :
Let's say that there are 6 CARDS A B C D E F; TWO SUITS - HEARTS AND SPADES
HEARTS: A B C D E F
SPADES: A B C D E F
Method 1 :
The first card could be chosen in 12C1/12C1 ways.
Second card - 1C1/11C1 (Because, there are only two same valued cards)
Third Card - 10C1/10C1
Fourth card - 8C1/9C1
Now, combinations of AABC = 4!/ (2!) = 12
Therefore, probability =
12x1x10x8
=---------- X 12 = 32/33 which is incorrect.
12x11x10x9
Method 2:
I tried another method:
We can choose one number from 1 to 6 in 6C1 ways. If we have to choose another number for pair, combinations = 6C1X 1C1 (Because if we choose, say, "2", then there is only one more number left to choose. Therefore, 1C1)
we can then choose 1 number out of the remaining 5 numbers in 5C1 X 2 (there are two sets of 5 such numbers)
Similarly, the last number can be chosen in 4C1 X 2 ways.
Therefore, probability =
6c1 x 1c1 x (5c1 x 2 ) x (4c1 x 2)
= ------------------------------------
12c4
= 32/33 which is again incorrect.
Can someone please help me? I got the same answer using two different methods, both of which are wrong
Thanks
MGMAT probability - need expert help
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You did a mistake of considering only 1 pair of set.
Whenever you find a probability question where there could be numerous combinations, always go for the negation approach.
Prob. to ger atleast 1 pair = 1 - Prob. to get no pair
Now Prob. to get no pair:
Prob. to get 1st card = 12/12
Prob. to get 2nd card with no pair = 10/11
Prob. to get 3rd card with no pair = 8/10
Prob. to get 3rd card with no pair = 6/9
=>Prob. to get no pair = 12/12 * 10/11 * 8/10 * 6/9 = 16/33
Prob. to ger atleast 1 pair = 1 - 16/33 = 17/33
Please recheck OA should be C).
Hope it helps
Whenever you find a probability question where there could be numerous combinations, always go for the negation approach.
Prob. to ger atleast 1 pair = 1 - Prob. to get no pair
Now Prob. to get no pair:
Prob. to get 1st card = 12/12
Prob. to get 2nd card with no pair = 10/11
Prob. to get 3rd card with no pair = 8/10
Prob. to get 3rd card with no pair = 6/9
=>Prob. to get no pair = 12/12 * 10/11 * 8/10 * 6/9 = 16/33
Prob. to ger atleast 1 pair = 1 - 16/33 = 17/33
Please recheck OA should be C).
Hope it helps
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Please re-read the question. The prompt has asked us to find the probability to find a pair and NOT AT LEAST ONE PAIR! OA is correct. I have verified the answer - it's 16/33.Prob. to ger atleast 1 pair = 1 - 16/33 = 17/33
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Oh, Sorry.
Here is the solution in that case:
Method 1:
We may have 6 combinations in the case of a pair = 4C2 = 6 (not 12)
1st selected number has a pair: AABC , ABAC, ABCA
2nd selected number has a pair: ABBC, ABCB
3rd selected number has a pair: ABCC
1 case has number of ways = 12/12*1/11* 10/10* 8/9 = 8/99
Total number of ways = 6 * 8/99 = 16/33
Method2:
Here you are taking 2 sets of 6 numbers each with same values. Different ways are:
A) 1 number from set 1, 3 numbers from set 2 = (6C1) * (1C1 * 5C2) = 60
B) 2 numbers from set 1, 2 numbers from set 2 = (6C2) * (2C1 * 4C1) = 120 (Out of 2 numbers one should match)
C) 3 numbers from set 1, 1 number from set 2 = (6C3) * (3C1) = 60 (Out of 3 numbers one should match)
Total number of required ways = 60 + 120 + 60 = 240
Total number of ways = 12C4 = 495
Number of ways for 1 pair = 240/495 = 15 * 16 / 15 * 33 = 16/33
Method3:
As I already said, number of ways for atleast one pair = 17/33
Ways to get both pairs = 12/12 * 1/11 * 10/10 * 1/9 * 3 = 1/33 ( 3 ways: AABB, ABAB, ABBA)
So Number of ways to get 1 pair = 17/33 - 1/33 = 16/33
Hope it helps...
Here is the solution in that case:
Method 1:
We may have 6 combinations in the case of a pair = 4C2 = 6 (not 12)
1st selected number has a pair: AABC , ABAC, ABCA
2nd selected number has a pair: ABBC, ABCB
3rd selected number has a pair: ABCC
1 case has number of ways = 12/12*1/11* 10/10* 8/9 = 8/99
Total number of ways = 6 * 8/99 = 16/33
Method2:
Here you are taking 2 sets of 6 numbers each with same values. Different ways are:
A) 1 number from set 1, 3 numbers from set 2 = (6C1) * (1C1 * 5C2) = 60
B) 2 numbers from set 1, 2 numbers from set 2 = (6C2) * (2C1 * 4C1) = 120 (Out of 2 numbers one should match)
C) 3 numbers from set 1, 1 number from set 2 = (6C3) * (3C1) = 60 (Out of 3 numbers one should match)
Total number of required ways = 60 + 120 + 60 = 240
Total number of ways = 12C4 = 495
Number of ways for 1 pair = 240/495 = 15 * 16 / 15 * 33 = 16/33
Method3:
As I already said, number of ways for atleast one pair = 17/33
Ways to get both pairs = 12/12 * 1/11 * 10/10 * 1/9 * 3 = 1/33 ( 3 ways: AABB, ABAB, ABBA)
So Number of ways to get 1 pair = 17/33 - 1/33 = 16/33
Hope it helps...
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Why? I am not sure. Let's say that the cards are AABC. Total number of combinations = 4!/2!=12 i.e.MakeUrTimeCount wrote:Oh, Sorry.
Here is the solution in that case:
Method 1:
We may have 6 combinations in the case of a pair = 4C2 = 6 (not 12)
A A B C
A A C B
A _ A _ (2 each)
A _ _ A (2 each)
_ A A _ (2 each)
_ _ A A (2 each)
_ A _ A (2 each)
=12 combinations. Am I missing anything?
Thanks
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Yes, you are taking order into account. I'm not sure whether you play poker, but when you look at your whole cards (in Texas Hold 'Em) and see a pair of aces, it doesn't matter which is on the left and which is on the right as you look at them.voodoo_child wrote:Why? I am not sure. Let's say that the cards are AABC. Total number of combinations = 4!/2!=12 i.e.MakeUrTimeCount wrote:Oh, Sorry.
Here is the solution in that case:
Method 1:
We may have 6 combinations in the case of a pair = 4C2 = 6 (not 12)
A A B C
A A C B
A _ A _ (2 each)
A _ _ A (2 each)
_ A A _ (2 each)
_ _ A A (2 each)
_ A _ A (2 each)
=12 combinations. Am I missing anything?
Thanks
This is a pretty basic statistical concept: always determine whether order matters. If it does, then you need to count all possibilities. If it doesn't, then you will always end up with a smaller number, because you need to eliminate the duplicates.
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Thanks Krusta80 and everyone who shared thoughts. Any tips on how to find it ? For instance, the following example assumes that the order counts:krusta80 wrote:Yes, you are taking order into account. I'm not sure whether you play poker, but when you look at your whole cards (in Texas Hold 'Em) and see a pair of aces, it doesn't matter which is on the left and which is on the right as you look at them.voodoo_child wrote:Why? I am not sure. Let's say that the cards are AABC. Total number of combinations = 4!/2!=12 i.e.MakeUrTimeCount wrote:Oh, Sorry.
Here is the solution in that case:
Method 1:
We may have 6 combinations in the case of a pair = 4C2 = 6 (not 12)
A A B C
A A C B
A _ A _ (2 each)
A _ _ A (2 each)
_ A A _ (2 each)
_ _ A A (2 each)
_ A _ A (2 each)
=12 combinations. Am I missing anything?
Thanks
This is a pretty basic statistical concept: always determine whether order matters. If it does, then you need to count all possibilities. If it doesn't, then you will always end up with a smaller number, because you need to eliminate the duplicates.
In how many different ways can the letters A,A,B,B,B,C,D be arranged if the letter C must be to the right of the letter D?
The solution is straightforward. However, there will be a huge difference in the answer if we consider order and "no-order"!
Any tips?
Thanks
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"Arrangement" is one of those key words that indicates order mattering. Also, the second sentence helps confirm that order matters, since it describes the location of one letter versus another. Again, location means order matters means permutations.voodoo_child wrote:Thanks Krusta80 and everyone who shared thoughts. Any tips on how to find it ? For instance, the following example assumes that the order counts:krusta80 wrote:Yes, you are taking order into account. I'm not sure whether you play poker, but when you look at your whole cards (in Texas Hold 'Em) and see a pair of aces, it doesn't matter which is on the left and which is on the right as you look at them.voodoo_child wrote:Why? I am not sure. Let's say that the cards are AABC. Total number of combinations = 4!/2!=12 i.e.MakeUrTimeCount wrote:Oh, Sorry.
Here is the solution in that case:
Method 1:
We may have 6 combinations in the case of a pair = 4C2 = 6 (not 12)
A A B C
A A C B
A _ A _ (2 each)
A _ _ A (2 each)
_ A A _ (2 each)
_ _ A A (2 each)
_ A _ A (2 each)
=12 combinations. Am I missing anything?
Thanks
This is a pretty basic statistical concept: always determine whether order matters. If it does, then you need to count all possibilities. If it doesn't, then you will always end up with a smaller number, because you need to eliminate the duplicates.
In how many different ways can the letters A,A,B,B,B,C,D be arranged if the letter C must be to the right of the letter D?
The solution is straightforward. However, there will be a huge difference in the answer if we consider order and "no-order"!
Any tips?
Thanks
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While reviewing this problem, I thought of finding the probability of finding 1 pair using combinations. I know that we have already discussed one method above. I want to try another one.
Set1 - A B C D E F
Set2 - A B C D E F
The first card could be chosen in 12C1 ways.
Second card - 1C1 (Because, there are only two same valued cards)
Third Card - 10C1
Fourth card - 8C1
Now, combinations of AABC = 4!/ (2!) = 12
There are six possible arrangements for AABC, as discussed above
Therefore, probability =
12x1x10x8
=---------- X 6 = 64/33 = > Incorrect.
12C4
Any thoughts why?>
Thanks
Set1 - A B C D E F
Set2 - A B C D E F
The first card could be chosen in 12C1 ways.
Second card - 1C1 (Because, there are only two same valued cards)
Third Card - 10C1
Fourth card - 8C1
Now, combinations of AABC = 4!/ (2!) = 12
There are six possible arrangements for AABC, as discussed above
Therefore, probability =
12x1x10x8
=---------- X 6 = 64/33 = > Incorrect.
12C4
Any thoughts why?>
Thanks
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The total number of ways to pick 4 cards from 12 is 12C4 = 12!(4!8!) = 495.
To make exactly one pair, we need two matching cards and 2 other cards that don't match. There are 6 ways to make a pair: 1-1, 2-2, etc. For the other cards, we are choosing 2 from the remaining 10: 10C2 = 10!/(2!8!) = 45. However, within those 45 combinations we have the other 5 pairs we can make, which would give us 2 pairs. To avoid these, we subtract those 5 from 45 for a total of 40 ways to draw the final two cards without making a pair.
6 pairs * 40 non-pairs = 240 ways to draw 4 cards and get exactly one pair.
P=240/495 = 16/33
To make exactly one pair, we need two matching cards and 2 other cards that don't match. There are 6 ways to make a pair: 1-1, 2-2, etc. For the other cards, we are choosing 2 from the remaining 10: 10C2 = 10!/(2!8!) = 45. However, within those 45 combinations we have the other 5 pairs we can make, which would give us 2 pairs. To avoid these, we subtract those 5 from 45 for a total of 40 ways to draw the final two cards without making a pair.
6 pairs * 40 non-pairs = 240 ways to draw 4 cards and get exactly one pair.
P=240/495 = 16/33
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Bill at VeritasPrep already gave a very good explanation, but because you sent me a pm, I will write a bit here as well.voodoo_child wrote:While reviewing this problem, I thought of finding the probability of finding 1 pair using combinations. I know that we have already discussed one method above. I want to try another one.
Set1 - A B C D E F
Set2 - A B C D E F
The first card could be chosen in 12C1 ways.
Second card - 1C1 (Because, there are only two same valued cards)
Third Card - 10C1
Fourth card - 8C1
Now, combinations of AABC = 4!/ (2!) = 12
There are six possible arrangements for AABC, as discussed above
Therefore, probability =
12x1x10x8
=---------- X 6 = 64/33 = > Incorrect.
12C4
Any thoughts why?
First of all, the difference between permutations vs. combinations is crucial. In permutations, order matters. In combinations, order doesn't matter. When Bill plays this game with Tracy's deck, the order doesn't matter --- it doesn't matter whether the two cards of the pair are picked first and second, or second and fourth, or whatever. BUT, the way you are analyzing --- First card picked = first card of pair, second card picked = second card of pair, etc. ---- that implicitly introduced an ordering requirement that is not part of the question. That's reason #1 your approach didn't work.
This type of question --- pick a few items, and some of them meet a special condition and some don't --- these are notoriously difficult questions in combinatorics. Don't feel bad if your find questions of this genre challenging -- they are challenging!
The way to approach this is as Bill did.
First, calculate the total number of combinations --- what in statistics/probability theory is called the "sample space". Here, that's 12C4 = 495, the total number of four-card combinations that could arise from Tracy's 12-card deck.
Now, of those, 495 combinations, how many have one pair only?
The number of possible pairs is six (AA, BB, etc.)
With any given pair, how many ways can the other two cards be different?
One way to answer this question (the route Bill took) is to say: let's take all possible combinations of two cards from the remaining ten ---- 10C2 = 45 --- and then subtract the other five pairs that could arise, because we want only one pair. Thus 45 - 5 = 40
Another way is to say -- OK, we already have the pair. Now, how many ways to pick the third card? Ten ways. Then, how many ways to pick a fourth card that is different from both the pair and the third card? Eight ways. BUT, notice we have counted, for example, both AABC and AACB as if they were two different things, and they are not. Therefore, once we multiply 8*10 = 80, we need to divide by 2 so we are not counting the same thing twice. That's 80/2 = 40. Two ways to get the same answer.
Six possible pairs, each with 40 possible non-matching pairs accompanying them = 240 combinations with exactly one pair.
240/495 = 48/99 = 16/33
Does all that make sense?
Please let me know if you have further questions.
Mike
Last edited by Mike@Magoosh on Sun May 06, 2012 11:34 am, edited 1 time in total.
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Thanks Mike! Perfect explanation. I see my mistake now.Mike@Magoosh wrote:Bill at VeritasPrep already gave a very good explanation, but because you sent me a pm, I will write a bit here as well.voodoo_child wrote:While reviewing this problem, I thought of finding the probability of finding 1 pair using combinations. I know that we have already discussed one method above. I want to try another one.
Set1 - A B C D E F
Set2 - A B C D E F
The first card could be chosen in 12C1 ways.
Second card - 1C1 (Because, there are only two same valued cards)
Third Card - 10C1
Fourth card - 8C1
Now, combinations of AABC = 4!/ (2!) = 12
There are six possible arrangements for AABC, as discussed above
Therefore, probability =
12x1x10x8
=---------- X 6 = 64/33 = > Incorrect.
12C4
Any thoughts why?
First of all, the difference between permutations vs. combinations is crucial. In permutations, order matters. In combinations, order doesn't matter. When Bill plays this game with Tracy's deck, the order doesn't matter --- it doesn't matter whether the two cards of the pair are picked first and second, or second and fourth, or whatever. BUT, the way you are analyzing --- First card picked = first card of pair, second card picked = second card of pair, etc. ---- that implicitly introduced an ordering requirement that is not part of the question. That's reason #1 your approach didn't work.
This type of question --- pick a few items, and some of them meet a special condition and some don't --- these are notoriously difficult questions in combinatorics. Don't feel bad if your find questions of this genre challenging -- they are challenging!
The way to approach this is as Bill did.
First, calculate the total number of combinations --- what in statistics/probability theory is called the "sample space". Here, that's 12C4 = 495, the total number of four-card combinations that could arise from Tracy's 12-card deck.
Now, of those, 495 combinations, how many have one pair only?
The number of possible pairs is six (AA, BB, etc.)
With any given pair, how many ways can the other two cards be different?
One way to answer this question (the route Bill took) is to say: let's take all possible combinations of two cards from the remaining ten ---- 10C2 = 45 --- and then subtract the other five pairs that could arise, because we want only one pair. Thus 45 - 5 = 40
Another way is to say -- OK, we already have the pair. Now, how many ways to pick the third card? Ten ways. Then, how many ways to pick a fourth card that is different from both the pair and the third card? Eight ways. BUT, notice we have counted, for example, both AABC and AACB as if they were two different things, and they are not. Therefore, once we multiply 8*10 = 80, we need to divide by 2 so we are not counting the same thing twice. That's 80/2 = 40. Two ways to get the same answer.
Six possible pairs, each with 40 possible non-matching pairs accompanying them = 240 combinations with exactly one pair.
240/495 = 48/99 = 16/33
Does all that make sense?
You might find this intro video about combinatorics helpful:
https://gmat.magoosh.com/lessons/334-lis ... d-counting
Please let me know if you have further questions.
Mike
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Mike, While reviewing this problem, I thought of repeating the problem with another Combinations method.Mike@Magoosh wrote:
Please let me know if you have further questions.
Mike
I took a subset of cards (say)
Set 1 - A B C
Set 2 - A B C
Now, there are 6C2=15 combinations to pick *any* two cards. Out of 15 cards, 3 are pairs. Out of 12 combinations, A{chosen from set1}A{chosen from set2} = A{chosen from set2} A{chosen from set1}.
Therefore, we need to divide by 2 => (12-3)=6. Similarly, I will have pairs such as AB=BA; hence, we need to again divide by 2 => 6/2 = 3.
Therefore, there are 3 pairs, 3 non-pairs.
Using the same concept, 12C2= 66. 6 out of 66 are pairs. 60/4=15.
Therefore, number of ways to choose a pair = 6C1 * 15C1 (i.e. choose one pair from 6 possible pairs and choose 1 pair from the remaining 15 non-pairs)
However, 15*6 = 90.
Can you please let me know the error in calculation above? Your response is greatly appreciated.
Thanks
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Dear voodoo_child
Yes, there's a big problem in what you did . . .
When you are applying the Fundamental Counting Principle (FCP) (i.e. "this many possibilities for the first choice, this many possibilities for the second choice, etc.), then you have to divide by 2 (or whatever higher nCk) to account for duplicates --- that system will count both AB and BA, and you have to account for that.
By contrast, when you use nCk, that automatically accounts for duplicates already --- they are already not counted, so the number you get, nCk, is the real number of combination of k elements chosen from a pool of n. If AB is counted, BA automatically will not be counted, just by the way nCk is set up.
When you calculate 6C2 = 15, then there actually are fifteen distinct pairs, no duplicates. As you note, 3 of them are two-of-a-kind pairs, so they are 15 - 3 = 12 distinct non-matching pairs. There are 3 matching pairs and 12 non-matching pairs.
For clarity, I'll list them. Let's say
Set 1 = a b c
Set 2 = A B C
3 matching pairs = Aa, Bb, Cc
12 non-matching pairs = Ab, AB, Ac, AC, ab, ac, Ba, Bc, BC, bc, Ca, Cb
Probability of picking a matching pair from this set
= (number of matching pairs)/(total number of pairs) = 3/15 = 1/5
===================================
Now, move to
Set 1 = A B C D E F
Set 2 = a b c d e f
Now there are 12C2 = 66 pairs. There really are 66 unique and different pairs, no need to divide anything by two. Of those pairs, 6 are matching pairs, so the other 60 are non-matching pairs.
Probability of picking a matching pair from this set on a two card draw
= (number of matching pairs)/(total number of pairs) = 6/66 = 1/11
The harder question, the original question --- what is the probability of choosing four cards and having exactly two of them match as a pair?
You can't just multiply the total number of matching pairs and the total number of non-matching pairs --- among other things, some of those matching pairs (e.g. Bb) will overlap with some of the possible non-matching pairs (e.g. Bf or Db). With any given matching pair, only certain non-matching pairs are possible.
There are still six pairs. To calculate the number of non-matching pairs that can go with any particular pair, you have to, as it were, set aside that pair and calculate anew with the remaining 10 cards. There are two ways to do that.
1) using nCk
Set aside the matching pair. Of the remaining 10 cards, there are 10C2 = 45 pairs. There really are 45 pairs, no need to divide anything by 2. Of those 45 pairs, 5 form a second matching pair, which we don't want, so there are 40 non-matching pairs.
2) using FCP
Set aside the matching pair. Of the remaining 10 cards, there are 10 choices for the third card, and if we don't want a match, there are 8 choices for the fourth man. Because we are using the FCP, and not nCk, that means we have counted pairs twice --- eF and Fe have both been counted. Therefore, the 8*10 = 80 needs to be divided by 2 --- and again, we wind up with 40 non-matching pairs.
6 matching pairs, each with 40 non-matching pairs that can go with it ==> 240 combinations meet the condition.
That's of all possible sets of 4 ===> 12C4 = 495
So, probability = 240/495 = 80/165 = 16/33
Does all this make sense? Do you see the difference between calculating numbers with the FCP vs. with nCk? Please let me know if have any further questions.
Mike
Yes, there's a big problem in what you did . . .
When you are applying the Fundamental Counting Principle (FCP) (i.e. "this many possibilities for the first choice, this many possibilities for the second choice, etc.), then you have to divide by 2 (or whatever higher nCk) to account for duplicates --- that system will count both AB and BA, and you have to account for that.
By contrast, when you use nCk, that automatically accounts for duplicates already --- they are already not counted, so the number you get, nCk, is the real number of combination of k elements chosen from a pool of n. If AB is counted, BA automatically will not be counted, just by the way nCk is set up.
When you calculate 6C2 = 15, then there actually are fifteen distinct pairs, no duplicates. As you note, 3 of them are two-of-a-kind pairs, so they are 15 - 3 = 12 distinct non-matching pairs. There are 3 matching pairs and 12 non-matching pairs.
For clarity, I'll list them. Let's say
Set 1 = a b c
Set 2 = A B C
3 matching pairs = Aa, Bb, Cc
12 non-matching pairs = Ab, AB, Ac, AC, ab, ac, Ba, Bc, BC, bc, Ca, Cb
Probability of picking a matching pair from this set
= (number of matching pairs)/(total number of pairs) = 3/15 = 1/5
===================================
Now, move to
Set 1 = A B C D E F
Set 2 = a b c d e f
Now there are 12C2 = 66 pairs. There really are 66 unique and different pairs, no need to divide anything by two. Of those pairs, 6 are matching pairs, so the other 60 are non-matching pairs.
Probability of picking a matching pair from this set on a two card draw
= (number of matching pairs)/(total number of pairs) = 6/66 = 1/11
The harder question, the original question --- what is the probability of choosing four cards and having exactly two of them match as a pair?
You can't just multiply the total number of matching pairs and the total number of non-matching pairs --- among other things, some of those matching pairs (e.g. Bb) will overlap with some of the possible non-matching pairs (e.g. Bf or Db). With any given matching pair, only certain non-matching pairs are possible.
There are still six pairs. To calculate the number of non-matching pairs that can go with any particular pair, you have to, as it were, set aside that pair and calculate anew with the remaining 10 cards. There are two ways to do that.
1) using nCk
Set aside the matching pair. Of the remaining 10 cards, there are 10C2 = 45 pairs. There really are 45 pairs, no need to divide anything by 2. Of those 45 pairs, 5 form a second matching pair, which we don't want, so there are 40 non-matching pairs.
2) using FCP
Set aside the matching pair. Of the remaining 10 cards, there are 10 choices for the third card, and if we don't want a match, there are 8 choices for the fourth man. Because we are using the FCP, and not nCk, that means we have counted pairs twice --- eF and Fe have both been counted. Therefore, the 8*10 = 80 needs to be divided by 2 --- and again, we wind up with 40 non-matching pairs.
6 matching pairs, each with 40 non-matching pairs that can go with it ==> 240 combinations meet the condition.
That's of all possible sets of 4 ===> 12C4 = 495
So, probability = 240/495 = 80/165 = 16/33
Does all this make sense? Do you see the difference between calculating numbers with the FCP vs. with nCk? Please let me know if have any further questions.
Mike
Magoosh GMAT Instructor
https://gmat.magoosh.com/
https://gmat.magoosh.com/