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Bill has a set of 6 black cards and a set of 6 red cards. Each card 08 Apr 2021, 00:32
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dave13
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VeritasKarishma thanks, is it possible that you use my method for solving question ? :)

That method involves multiple levels of calculations which makes it cumbersome at test time.

A pair and two Reds:
6 * 5C2 = 6 * 10 = 60

A pair and two Blacks:
6 * 5C2 = 6 * 10 = 60

A pair and a red and a black non pair:
6 * 5 * 4 = 120
5 ways to pick a red and 4 ways to pick a non matching black.

(All I have done in my method above is combined these three steps together to get 240)

Two pairs:
6C2 = 15

Total = 255


many thanks VeritasKarishma , i almost understand it :) just have a few questions to make it clear for me :grin:

as per you method (see highlighted portion below):

"Select 1 pair out of the 6 in 6C1 ways = 6 ways (say you selected (R1, B1))
Of the leftover 10 cards, select any two in 10C2 = 45 ways.
But of these 45 ways, 5 are such that you have another pair (R2, B2), (R3, B3) ... (R6, B6)
So you have 40 ways of selecting exactly 1 pair.
Total = 6 * 40 = 240 ways"


you say that out of 10 cards, you still can have pairs so to avoid another same pair you exclude 5 cards (R2, B2), (R3, B3) ... (R6, B6)

and you are left with 40 ways ... this is how i understand it ...though a bit confusing

but when i wrote this:

" - 1R1B2R3B choosing one red and one black of the same value and the rest two are one black and one red of different values = 6*1*5*3 = 90 (excluded 4 to avoid the second pair to be the same) "

here i excluded 4 to avoid the second pair to be the same.... but you say its not correct. is it not correct at all, or just in this case? in which would it be correct ? do i arrange here too ? i m curious to know :)


Btw, here "A pair and a red and a black non pair: 6 * 5 * 4 = 120" if third card is red which is fifth, in the rest of 4 cards there is a card of identical value among black ones left, so my question is why you didnt exclude that black card .... i mean i would do it 6*5*3 .... thats kinda confuses me
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Yes, out of 10 when you pick 2 cards, they can be anything (R2, B3), (R4, R5), (B2, B6)... etc. You get 45 such cases. But 5 of these cases will be (R2, B2), (R3, B3), (R4, B4), (R5, B5), (R6, B6)
which give us another pair. So you need to remove them.

Quote:

here i excluded 4 to avoid the second pair to be the same.... but you say its not correct. is it not correct at all, or just in this case? in which would it be correct ? do i arrange here too ? i m curious to know :)
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I don't understand what you mean by - "I excluded 4".

If you want 1 pair and 1 red, 1 black of different values, you will select these as:

6C1 (6 ways to select a pair. Say you got R1, B1)

Now there are 5 ways to select R (R2, R3, R4, R5, R6). Say you got R2.
Now there are 4 ways to select B (B3, B4, B5, B6) to ensure that you get different values. You have already removed B2 to ensure that you do not get a pair. You are still left with 4 different values of B.

So it will be 6 * 5 * 4
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VeritasKarishma thanks, i now tried to approach this problem with probability method (but not using 1-all non desirable outcomes) just regular ...

so i did this


case 1: one pair is the same

12/12 * 1/11 * 10/10 * 9/9 = 1/11 *2 =2/11 (since it can be 11RR or 11BB i multiplied by two)

case 2: two pairs are same

12/12 * 1/11* 10/10 *9/9 = 1/99

1/99+ 2/11 = 19/99

whats wrong with my approach ?
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VeritasKarishma thanks, i now tried to approach this problem with probability method (but not using 1-all non desirable outcomes) just regular ...

so i did this


case 1: one pair is the same

12/12 * 1/11 * 10/10 * 9/9 = 1/11 *2 =2/11 (since it can be 11RR or 11BB i multiplied by two)

case 2: two pairs are same

12/12 * 1/11* 10/10 *9/9 = 1/99

1/99+ 2/11 = 19/99

whats wrong with my approach ?
Show more

You have not arranged them. But arranging these probabilities is a bit tricky since you need to know the cases that you have covered in your probability and those that you haven't.

For example: The case of 2 pairs

1 * (1/11) * 1 * (1/9)

Pick anything * Pick its pair * Pick anything else * Pick its pair

Say the two pairs you pick are (3, 3, 4, 4). So you have already taken (4, 4, 3, 3) into account too because when you pick anything first, you could have picked 4 first too. But you haven't taken (3, 4, 3, 4) and (3, 4, 4, 3) into account.
So the entire thing gets multiples by 3 to make up for these cases.

Hence, the better approach is combinatorics or that in which arrangement is irrelevant (e.g. when you pick no pair - pick anything and follow it up by picking anything other than what you have already picked in each turn)
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VeritasKarishma thanks, i now tried to approach this problem with probability method (but not using 1-all non desirable outcomes) just regular ...

so i did this


case 1: one pair is the same

12/12 * 1/11 * 10/10 * 9/9 = 1/11 *2 =2/11 (since it can be 11RR or 11BB i multiplied by two)

case 2: two pairs are same

12/12 * 1/11* 10/10 *9/9 = 1/99

1/99+ 2/11 = 19/99

whats wrong with my approach ?

You have not arranged them. But arranging these probabilities is a bit tricky since you need to know the cases that you have covered in your probability and those that you haven't.

For example: The case of 2 pairs

1 * (1/11) * 1 * (1/9)

Pick anything * Pick its pair * Pick anything else * Pick its pair

Say the two pairs you pick are (3, 3, 4, 4). So you have already taken (4, 4, 3, 3) into account too because when you pick anything first, you could have picked 4 first too. But you haven't taken (3, 4, 3, 4) and (3, 4, 4, 3) into account.
So the entire thing gets multiples by 3 to make up for these cases.

Hence, the better approach is combinatorics or that in which arrangement is irrelevant (e.g. when you pick no pair - pick anything and follow it up by picking anything other than what you have already picked in each turn)
Show more


VeritasKarishma as per the same logic as case of two pairs the same ... i did this

for one pair the same = 12/12 * 1/11 *10/10 * 8/9 = 8/99

since there can be arrangements such as 3345 , 4533, 3453, 3435 and 4353 multiply 8/99 by 5

so i get 40/99

1/33+40/99 = 43/99

now whats wrong ? :?

update: i googled this question and for one pair the same found this solution (12/12)(1/11)(10/10)(8/9) * (4!/2!2!)


why here we are divide 4! by 2! 2! ? if for example i have this arrangement 1134 then total number of arrangements is 4!/2!

curious to know :)
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hello :) GMATGuruNY IanStewart AndrewN

"Probablity of selecting one pair :
(12/12) * (1/12)* 10/10 *8/9= 8/99----assuming ist two cards are same.
12/12 * 10/11* 2/10 * 8/9 = 16/99 -------assuming 1st and third card are same .
12/12 * 10/11 * 8 /10 * 3/9 = 24/99-------assuming ist and fourth card are same.
Total probablity of only one card same is therefore : (8+16+24)/99 = 48/99 =16/33-----(1)
Probablity of two pair same in similar way is
12/12 *1/12* 10/10*1/9 = 1/99 ....ist two same and third and fourth same.
12/12*10/11*2/10*1/9 = 2/99----- This includes both 1-3 * 2-4 similar and 1-4 & 2-3 similar
Therefore total probablity of any two same is 3/99=1/33----(2)

therefore total probablity = (1) + (2) = 17/33. "


in the solution (not by me) above

the case when one pair is the same,

there are there scenarios

First Scenario: 1134

Second Scenario: 1314

Third Scenario: 1341


i understand that 1314 is the same as 1413, similarly 1341 is the same as 1431


But there are ALSO scenarios such as 3411 and 3141 these are two distinct scenarios that are not counted via probability method above, Why?

to elaborate further i ask myself question :)

but in the solution (not by me) below for case 1

(12/12)(1/11)(10/10)(8/9) * (4!/2!2!)

the result is multiplied by (4!/2!2!)

from here i can deduce that since among four letters (1134) two are the same one must divide by 2! Also since 1314 is the same as 1413 one should divide by another 2!

BUT (4!/2!2!) yields 6 total arrangements

Question 1: why when using probability we use only 3 scenarios whereas there are total SIX scenarios/arrangements ?

Question 2: (4!/2!2!) yields 6 total arrangements , What are these 6 total arrangements ? i can only see five arrangements: 1134, 1314, 1341, 3411, 3141

thank you! :grin: :)
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dave13
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VeritasKarishma thanks, i now tried to approach this problem with probability method (but not using 1-all non desirable outcomes) just regular ...

so i did this


case 1: one pair is the same

12/12 * 1/11 * 10/10 * 9/9 = 1/11 *2 =2/11 (since it can be 11RR or 11BB i multiplied by two)

case 2: two pairs are same

12/12 * 1/11* 10/10 *9/9 = 1/99

1/99+ 2/11 = 19/99

whats wrong with my approach ?

You have not arranged them. But arranging these probabilities is a bit tricky since you need to know the cases that you have covered in your probability and those that you haven't.

For example: The case of 2 pairs

1 * (1/11) * 1 * (1/9)

Pick anything * Pick its pair * Pick anything else * Pick its pair

Say the two pairs you pick are (3, 3, 4, 4). So you have already taken (4, 4, 3, 3) into account too because when you pick anything first, you could have picked 4 first too. But you haven't taken (3, 4, 3, 4) and (3, 4, 4, 3) into account.
So the entire thing gets multiples by 3 to make up for these cases.

Hence, the better approach is combinatorics or that in which arrangement is irrelevant (e.g. when you pick no pair - pick anything and follow it up by picking anything other than what you have already picked in each turn)


VeritasKarishma as per the same logic as case of two pairs the same ... i did this

for one pair the same = 12/12 * 1/11 *10/10 * 8/9 = 8/99

since there can be arrangements such as 3345 , 4533, 3453, 3435 and 4353 multiply 8/99 by 5

so i get 40/99

1/33+40/99 = 43/99

now whats wrong ? :?

update: i googled this question and for one pair the same found this solution (12/12)(1/11)(10/10)(8/9) * (4!/2!2!)


why here we are divide 4! by 2! 2! ? if for example i have this arrangement 1134 then total number of arrangements is 4!/2!

curious to know :)
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Here are all the cases:

No Pair:

1 * (10/11) * (8/10) * (6/9) = 48/99
Pick any * Pick any except the one you picked previously * Pick any except two already picked * Pick any except the three already picked

This will give you all arrangements such 1234, 2134, 2341 etc.


Only one pair:

1 * (1/11) * 1 * (8/9)
Pick any * Pick the same again * Pick any * Pick any other

This will give you 1123, 1132, 2238, 2283.. etc
But, for 1123 we will not get 1213, 1231, 2113, 2131, 2311. So you need to multiply it by 6 (there are total 6 cases attached to each)

1 * (1/11) * 1 * (8/9) * 4!/2!*2! = 48/99
Why do we multiply by 4!/2!*2! (which is same as 6 we saw above)? Arrange all 4 digits in 4! ways. But two digits are same so divide by 2!. Also, distinct two digits have already been arranged (1123 and 1132 are different cases that we get because our third pick is also "pick any". So divide by 2! again.


Two pairs:

1 * (1/11) * 1 * (1/9)
Pick any * Pick its pair * Pick any * Pick its pair
This gives us 1122, 2211, 1155... etc

1 * (1/11) * 1 * (1/9) * 3 = 3/99
We multiply by 3 to account for all 3: 1122, 1212, 1221
Or we can say that we multiply by 4!/2!*2!*2! because 2 pairs are same also arrangement of smaller/greater pairs is already accounted for i.e. we count 1122 and 2211 as separate cases.
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dave13

"Probablity of selecting one pair :
(12/12) * (1/12)* 10/10 *8/9= 8/99----assuming ist two cards are same.
12/12 * 10/11* 2/10 * 8/9 = 16/99 -------assuming 1st and third card are same .
12/12 * 10/11 * 8 /10 * 3/9 = 24/99-------assuming ist and fourth card are same.

in the solution (not by me) above

But there are ALSO scenarios such as 3411 and 3141 these are two distinct scenarios that are not counted via probability method above, Why?
Show more

I'm not sure where that solution is from, but it's not explaining its cases correctly. The last case, for example, calculates the probability that it's the fourth card that makes the pair (with the first, second, or third card). It's not the probability that the "1st and fourth card are same". So the last case is considering all of the possibilities ABCA, BACA and BCAA. Similarly the second case covers two possibilities, when the first and third cards are pairs, and when the second and third cards are pairs.

dave13

Question 1: why when using probability we use only 3 scenarios whereas there are total SIX scenarios/arrangements ?
Show more

I don't think the solution you quote is the best one to follow -- there's no reason to even consider three scenarios here. Each of the six scenarios is equally likely, so you just need to consider one of them, then multiply by 6. The solution you quote collapses some of the scenarios into a single case (it makes one case of the three situations ABCA, BACA and BCAA, for example) which is why it has three instead of six cases. But solving this problem directly is longer than solving the opposite problem -- getting no pairs -- and subtracting from 1.

dave13

Question 2: (4!/2!2!) yields 6 total arrangements , What are these 6 total arrangements ? i can only see five arrangements: 1134, 1314, 1341, 3411, 3141
Show more

If you want to list arrangements in situations like this, it's best to proceed systematically, by changing only one thing at a time. I'd first list all of the arrangements with 'A' or '1' first position, then all of the additional ones where 'A' or '1' is in the second position, and so on. Here the six arrangements are:

AABC
ABAC
ABCA
BAAC
BACA
BCAA

In your notation, you missed 3114.
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hello :) GMATGuruNY IanStewart AndrewN
thank you! :grin: :)
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Hello, dave13. I see that both VeritasKarishma and IanStewart have answered your questions above. For my part, I solved the question using a hybrid method of logic and combinations. I knew that I had to get rid of the cases in which no pairs were selected, so I worked that out in the following manner:

The first card selected has 12 possibilities.

The second card selected has 10 possibilities, since it cannot match the first card selected.

The third card selected has 8 possibilities, since it cannot match either of the first two.

The fourth card selected has 6 possibilities, since it cannot match any of the first three.

Thus, the total number of ways to draw these cards would be

\(12*10*8*6\)

But I did not work that product out. Since the first card selected could just as easily have been the second, third, or fourth selected, we need to divide out the repetitions for each card selected. Thus,

\(\frac{12*10*8*6}{4*3*2*1}=240\)

There are 240 ways to draw these cards. We just need to know how many ways there are to draw 4 cards from 12. A quick 12C4 will get us that answer:

\(\frac{12!}{4!8!}=495\)

Finally, we need to figure out our desired cases and divide by the total we just derived.

\(\frac{495-240}{495}\)

\(\frac{255}{495}\)

Here, I divided the numerator and denominator by 10 to get more workable numbers, 25.5 and 49.5, and then multiplied by 2 to make them integers again. This gave me

\(\frac{51}{99}\)

I did not bother reducing it again. I just looked at the answers and chose (C), since 51 and 99 are both divisible by 3.

I am not advocating the above method over more efficient methods outlined earlier in the thread, but it can work in a pinch. (Sometimes I start working a problem one way before I realize I could have done it more efficiently in another.)

I hope this information is of some interest to you. Thank you for thinking to ask me about the question.

- Andrew
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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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@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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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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@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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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 14 Feb 2026, 05:23
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Hi Bunuel

I solve this question using probability method but I don't found answer

Probability of 1 pair= (12/12)*(1/11)*(10/10)*(8/9)*4!/2!*2!=(8/99)*6=16/33

Probability of 2 pair= (12/12)*(1/11)*(10/10)*(1/9)*4!/2!*2!= (1/99)*6=2/33

Answer is getting 18/33 instead of 17/33 where I am wrong


cumulonimbus
Probability Method:

Theory: Prob of at least 1 pair + Prob of 2 pairs

\(\frac{12}{12}*\frac{1}{11}*\frac{10}{10}*\frac{9}{9}\) * \(\frac{4!}{2!2!}\)

+

 \(\frac{12}{12}*\frac{1}{11}*\frac{10}{10}*\frac{1}{9}\) * \(\frac{4!}{2!2!}\) = 20/33.

Can you explain why r we multiplying by 4C2 ?

I get that we have 4 outcomes to be arranged. What is it divided by in the form of 2! * 2! ?
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