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705-805 Level|   Probability|      
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 08 Nov 2024, 23:29
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This reasoning is similar with mine

My combination: First select one group with similar number, that should be 6C1, then I can pick freely in the remaining 10 cards, which is 10C2. Last, I should minus the double-counted ones, the 2 pairs with the same number, which is 6C2.

So my mumerator is 6C1*10C2-6C2
ShashankDave
\frac{17}{33} is the answer.
You cant also do it using combinations.
select any pair(i.e. 2 cards) and the other two + select any two pairs and divide the total by 12C4.
this thought process leads to :-

(6C1*8*5 + 6C2)/12C4

for the first term in the Numerator - first select any one pair(6C1 ways), the we are left with 5 of black and 5 of red. Now select any one from one of these groups (5 ways). Lastly we are left with 9 cards, but in this group, one of them is the one with the same value in another color of the previously selected. So we are left with 8 to select one from(8 ways). Multiplying together -> 6C1*8*5.
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 15 Jul 2025, 11:55
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@Bunuel I'm a little confused here. Why is this not correct? 12C1*1*10C1*8C1 to pick one pair and two that are not pairs. (I pick one card from 12 and then only have one other card to pick so 1)*(I pick one from the remaining 10 cards)*(I pick 1 from the remaining 8 cards). Thank you!
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 15 Jul 2025, 12:42
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DrMock
@Bunuel I'm a little confused here. Why is this not correct? 12C1*1*10C1*8C1 to pick one pair and two that are not pairs. (I pick one card from 12 and then only have one other card to pick so 1)*(I pick one from the remaining 10 cards)*(I pick 1 from the remaining 8 cards). Thank you!
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Hi, a couple of errors in your approach,

1. We are looking for cases where we have at least 1 pair, so we will also need to include cases where we get two pairs; from your approach I assume you have incorrectly inferred that we only want one pair and the rest two should be different

2. If you're selecting cards one by one n times, you will also need to divide it by n! since you multiplied these selections and have arrangements of these selections which is not required

There are multiple correct approaches mentioned in the thread, would suggest you invest some time and go through them. Here's one good approach - https://gmatclub.com/forum/bill-has-a-s ... l#p1425104
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 15 Jul 2025, 13:10
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Hi Krunaal thanks for the response and clarification. On point 2, I had come to that conclusion as well. I guess I'm having trouble seeing the combination approach as NOT sequential. It was written somewhere else I that the numerator can be calculated by "Choose any 4 pair and then select any one from each pair. 6C4*2C1*2C1*2C1*2C1 = 240". Isn't this sequential in a sense as well? Thanks for your patience on this!
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DrMock
@Bunuel I'm a little confused here. Why is this not correct? 12C1*1*10C1*8C1 to pick one pair and two that are not pairs. (I pick one card from 12 and then only have one other card to pick so 1)*(I pick one from the remaining 10 cards)*(I pick 1 from the remaining 8 cards). Thank you!
Hi, a couple of errors in your approach,

1. We are looking for cases where we have at least 1 pair, so we will also need to include cases where we get two pairs; from your approach I assume you have incorrectly inferred that we only want one pair and the rest two should be different

2. If you're selecting cards one by one n times, you will also need to divide it by n! since you multiplied these selections and have arrangements of these selections which is not required

There are multiple correct approaches mentioned in the thread, would suggest you invest some time and go through them. Here's one good approach - https://gmatclub.com/forum/bill-has-a-s ... l#p1425104
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 15 Jul 2025, 13:25
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DrMock
Hi Krunaal thanks for the response and clarification. On point 2, I had come to that conclusion as well. I guess I'm having trouble seeing the combination approach as NOT sequential. It was written somewhere else I that the numerator can be calculated by "Choose any 4 pair and then select any one from each pair. 6C4*2C1*2C1*2C1*2C1 = 240". Isn't this sequential in a sense as well? Thanks for your patience on this!
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"Choose any 4 pair and then select any one from each pair" => this involves two steps, we first select the 4 numbers out of 6 by 6C4, now those 4 numbers can each be either red or black, so 2*2*2*2 => \(6C4*2^4\)

It may feel sequential because there are two steps, but we are not selecting individual cards in a particular order, we’re just building up the combination logically. Step 1 only narrows down which values will appear, and step 2 selects one card for each value.
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 16 Jul 2025, 00:26
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Krunaal Just for my better understanding, could you please help write out the numerator in a permutation then dividing it by n! cases where 1) I'm taking one pair and two non-pairs 2) Two pairs. I'm getting the right answer, like you said, when I take the opposite case, divide by n! and subtract it from one, but not when I take the other two scenarios. Thanks for your help.
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 07 Aug 2025, 10:09
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Hi, my approach to this was calculating ways in which I can choose 1 pair only + ways in which I can choose 2 pairs divided by ways in which I can choose 4 cards out of 12.

So essentially, [6C1 (for choosing that 1 pair) * 10C1 (for choosing 1 value from remaining 5 values, 2 colours each) * 8C1 (not choosing that 1 same value card) + 6C2 (for choosing 2 pairs out of 6)] / 12C4. Can you please help me understand where I am going wrong?
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