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Combination and Permutation Manhattan

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Combination and Permutation Manhattan

by Kaunteya » Thu Feb 28, 2008 9:13 am
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, 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 at least one pair of cards that have the same value?


a. 8/33

b. 62/165

c. 17/33

d. 103/165

e. 25/33

I got to there are 12C4 ways of picking four cards out of twelve, but what is the probabilty that of the four turned cards there are two with matching numbers? Help anyone. I really suck at these questions.

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by torontogmat.com » Thu Feb 28, 2008 2:17 pm
It does not matter which card is turned first.

There is a 1/11 chance that the next card turned will match. If it doesn't,
there is a 2/10 chance that the following card will match. If it still doesn't,
there is a 3/9 chance that the final card will match.

1/11 + 1/5 + 1/3 = 103 / 165.


I wish I could give some blanket statement about when to apply different methods of calculating probabilities. You add probabilities when it is enough for any event to be true, and you multiply them when they all must be true. However, there are a lot of questions when neither of these are the case.
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Combination and Permutation Manhattan

by jason49 » Mon Aug 04, 2008 6:30 am
I was looking for an answer to this question and came up with another thread that had a different answer, as reproduced below.

https://www.beatthegmat.com/probability- ... t5495.html

I think that C is correct, trying to figure out what tgmat calculated, because it obviously includes more cases than necessary.

Can somebody comment on this?


===========

I came up with C

Total 12 cards (1,2,3,4,5,6 and 1,2,3,4,5,6)
Ways to pick 4 cards (12*11*8*9)
Atleast one pair = 1 - (No pair)

How to pick a non pair ? For the first card we have 12 choices (assume we picked 6). For the second card we have 10 choices (cant pick 6. Assume the second card picked was 3). Third card we have 8 choices (Cant pick 6 and 3. Assume third card picked was 5). Fourth card we have 6 choices (Cant pick 6 ,3 or 5 )


So this boils down to = (12*10*8*6)/(12*11*8*9) = 16/33

Answer = 1-(16/33) = 17/33

HTH
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by sudhir3127 » Mon Aug 04, 2008 6:47 am
Even i go with C. 17/33

probability of atleast one = 1- p(0)

probability of not picking any of the maching cards in 4 draws is ..

1st draw is not important..

2 draw= we will have one card which is matching with the 1 drawn card .. so the probability of not picking up that card is 10/11

3 rd draw = now we will have 2 matching cards for the ones selected ... probabilty of not picking them is 8/10

4th draw = now we have 4 matching cards for the ones picked .. hence the probabilty is 6/9

thus the total probability of not picking any matching card is

10/11*8/10*6*9 =16/33

hence p( 1) = 1-p(0)

thus its 1- 16/33 = 17/33

Hope its clear...
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by Ian Stewart » Mon Aug 04, 2008 8:13 am
jason49 wrote: I think that C is correct, trying to figure out what tgmat calculated, because it obviously includes more cases than necessary.

Can somebody comment on this?
Nice find. In probability questions, when you see the words 'at least', as in this question, that often means that you would need to consider a few different cases if you did the problem directly. So this is often a clue that it will be faster to do the opposite problem: determine the probability that the event does not happen, then subtract from 1. That's exactly what sudhir did above, and that's exactly how I'd approach the problem as well.

That said, it is possible to do the problem directly, just as long as we break down the problem correctly. This solution is not quite right:
torontogmat.com wrote:It does not matter which card is turned first.

There is a 1/11 chance that the next card turned will match. If it doesn't,
there is a 2/10 chance that the following card will match. If it still doesn't,
there is a 3/9 chance that the final card will match.

1/11 + 1/5 + 1/3 = 103 / 165.
The probability, for example, that the last card matches one of the first three is not equal to 1/3, which is what the above solution assumes. It is only equal to 1/3 when we know that none of the first three cards match. We can adjust this solution to get the correct answer, however:

We get at least one pair IF:

[1+2 match] OR [(1+2 do not match)AND(1/2 matches 3)] OR [(1,2,3 do not match)AND(4 matches 1,2 or 3)]

The above solution missed the AND conditions. Solving:

1/11 + (10/11)(1/5) + [(10/11)(8/10)](1/3) = 15/165 + 30/165 + 40/165 = 85/165 = 17/33

Still, that's more difficult than the approach sudhir took above, which is much better.
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Re: Combination and Permutation Manhattan

by beeparoo » Mon Aug 04, 2008 1:00 pm
jason49 wrote:I was looking for an answer to this question and came up with another thread that had a different answer, as reproduced below.

https://www.beatthegmat.com/probability- ... t5495.html

I think that C is correct, trying to figure out what tgmat calculated, because it obviously includes more cases than necessary.

Can somebody comment on this?


===========

I came up with C

Total 12 cards (1,2,3,4,5,6 and 1,2,3,4,5,6)
Ways to pick 4 cards (12*11*10*9)
Atleast one pair = 1 - (No pair)

How to pick a non pair ? For the first card we have 12 choices (assume we picked 6). For the second card we have 10 choices (cant pick 6. Assume the second card picked was 3). Third card we have 8 choices (Cant pick 6 and 3. Assume third card picked was 5). Fourth card we have 6 choices (Cant pick 6 ,3 or 5 )


So this boils down to = (12*10*8*6)/(12*11*10*9) = 16/33

Answer = 1-(16/33) = 17/33

HTH
Correction in RED above.
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by CITI29 » Tue Aug 05, 2008 7:00 am
Hey guys,

this might be a very stupid question..but pls do clarify...that why in the above explanation we have used permutation 'Ways to pick 4 cards (12*11*10*9)' and not combination 12C4?...I dont undersatnd the logic of prob question very clearly so I just follow the basic pattern. So pls explain me this.

thanks.
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by beeparoo » Tue Aug 05, 2008 8:39 am
CITI29 wrote:Hey guys,

this might be a very stupid question..but pls do clarify...that why in the above explanation we have used permutation 'Ways to pick 4 cards (12*11*10*9)' and not combination 12C4?...I dont undersatnd the logic of prob question very clearly so I just follow the basic pattern. So pls explain me this.

thanks.
@CITI29: That is a good question that I've been scratching my head over as well.

If you read the other thread that someone (Jason49) linked above, someone explains that:
Because it doesn't say a group of 4 cards. Generally, whenever something is picked in a group, the question uses very explicit words (ex: group, team,committee,at once.. etc etc)
I'm not entirely convinced by this assessment, but it does offer me a little clue for spotting the nuance between permutation and combination.

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by Ian Stewart » Tue Aug 05, 2008 3:15 pm
CITI29 wrote: this might be a very stupid question..but pls do clarify...that why in the above explanation we have used permutation 'Ways to pick 4 cards (12*11*10*9)' and not combination 12C4?...I dont undersatnd the logic of prob question very clearly so I just follow the basic pattern. So pls explain me this.
Certainly not a stupid question. Remember that when computing probability we count two things: the number of ways we get the result we want, and the number of possible outcomes. It doesn't matter how you count these two things, as long as you're consistent. You can pretend order matters in both cases, or pretend order doesn't matter in both cases, and you shouldn't go wrong. Can't hurt to reiterate- just be sure you're consistent.

I'll give an example- say you're dealt two cards from a standard deck (52 cards, 4 of each type). If we want the probability that we get two Kings, we can pretend order does not matter:

There are 4C2 = 6 ways to get two kings (choosing two kings from four)
There are 52C2 = 52*51/2 ways to get two cards (choosing two cards from 52)

Divide, you get the probability: 6/(52*51/2) = 12/(52*51) = 1/(13*17)

Or you can pretend order matters- there are 4 ways the first card is a King, 3 ways the second: 4*3 = 12 ways in total.
There are 52*51 ways to get two cards, in order. 12/(52*51) = 1/(13*17)is the probability.

Both approaches give the same answer, of course. Normally, counting is much easier when order does matter, so most probability problems are more easily solved by pretending order matters, at least if you're counting possibilities in numerator and denominator. And of course you can do the above problem more simply, by using probability rules- I was just using it for illustration -

1/13 chance the first card is a King
3/51 chance the second is a King
(1/13)*(3/51) = 3/(13*51) = 1/(13*17), which is equal to both the probabilities we arrived at above.
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by beeparoo » Tue Aug 05, 2008 3:30 pm
WOW. :shock:

I had no idea until now that both techniques could be applied to the same problem so long as they were consistently used to achieve 1) the desired results and 2) the total possible results.

Ian, you just clarified something REALLY HUGE for me.

Thank you so very much!

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by CITI29 » Tue Aug 05, 2008 4:44 pm
THAAAANKS A TON!!!....:)
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by CITI29 » Tue Aug 05, 2008 4:44 pm
THAAAANKS A TON!!!....:)
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by tester_twelve » Sun Oct 03, 2010 5:26 pm
Ian Stewart wrote:
jason49 wrote: I think that C is correct, trying to figure out what tgmat calculated, because it obviously includes more cases than necessary.

Can somebody comment on this?
Nice find. In probability questions, when you see the words 'at least', as in this question, that often means that you would need to consider a few different cases if you did the problem directly. So this is often a clue that it will be faster to do the opposite problem: determine the probability that the event does not happen, then subtract from 1. That's exactly what sudhir did above, and that's exactly how I'd approach the problem as well.

That said, it is possible to do the problem directly, just as long as we break down the problem correctly. This solution is not quite right:
torontogmat.com wrote:It does not matter which card is turned first.

There is a 1/11 chance that the next card turned will match. If it doesn't,
there is a 2/10 chance that the following card will match. If it still doesn't,
there is a 3/9 chance that the final card will match.

1/11 + 1/5 + 1/3 = 103 / 165.
The probability, for example, that the last card matches one of the first three is not equal to 1/3, which is what the above solution assumes. It is only equal to 1/3 when we know that none of the first three cards match. We can adjust this solution to get the correct answer, however:

We get at least one pair IF:

[1+2 match] OR [(1+2 do not match)AND(1/2 matches 3)] OR [(1,2,3 do not match)AND(4 matches 1,2 or 3)]

The above solution missed the AND conditions. Solving:

1/11 + (10/11)(1/5) + [(10/11)(8/10)](1/3) = 15/165 + 30/165 + 40/165 = 85/165 = 17/33

Still, that's more difficult than the approach sudhir took above, which is much better.
Hi Ian/others,

Is there a way to solve this problem by first starting out with the total number of ways 4 cards can be drawn as the denominator (i.e. 12C4) and then determine what we're looking for in the numerator?
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by GMATGuruNY » Sun Oct 03, 2010 7:11 pm
tester_twelve wrote:
Ian Stewart wrote:
jason49 wrote: I think that C is correct, trying to figure out what tgmat calculated, because it obviously includes more cases than necessary.

Can somebody comment on this?
Nice find. In probability questions, when you see the words 'at least', as in this question, that often means that you would need to consider a few different cases if you did the problem directly. So this is often a clue that it will be faster to do the opposite problem: determine the probability that the event does not happen, then subtract from 1. That's exactly what sudhir did above, and that's exactly how I'd approach the problem as well.

That said, it is possible to do the problem directly, just as long as we break down the problem correctly. This solution is not quite right:
torontogmat.com wrote:It does not matter which card is turned first.

There is a 1/11 chance that the next card turned will match. If it doesn't,
there is a 2/10 chance that the following card will match. If it still doesn't,
there is a 3/9 chance that the final card will match.

1/11 + 1/5 + 1/3 = 103 / 165.
The probability, for example, that the last card matches one of the first three is not equal to 1/3, which is what the above solution assumes. It is only equal to 1/3 when we know that none of the first three cards match. We can adjust this solution to get the correct answer, however:

We get at least one pair IF:

[1+2 match] OR [(1+2 do not match)AND(1/2 matches 3)] OR [(1,2,3 do not match)AND(4 matches 1,2 or 3)]

The above solution missed the AND conditions. Solving:

1/11 + (10/11)(1/5) + [(10/11)(8/10)](1/3) = 15/165 + 30/165 + 40/165 = 85/165 = 17/33

Still, that's more difficult than the approach sudhir took above, which is much better.
Hi Ian/others,

Is there a way to solve this problem by first starting out with the total number of ways 4 cards can be drawn as the denominator (i.e. 12C4) and then determine what we're looking for in the numerator?
Yes, although it's not the approach that I would recommend.

Total number of ways to choose a combination of 4 cards from 12 choices = 12C4 = 495.

To get at least 1 pair of the same value, we need either to combine 1 pair of the same value with 1 pair of different values or to combine 2 pairs, each of which is of the same value.

Total number of ways to choose exactly 1 pair of the same value:
Number of ways to choose a pair of the same value = 6

Number of ways to choose a pair of different values from the 10 remaining cards:
1st card = 10 choices
2nd card = 8 choices (9 cards left, but we can't choose the mate of the card just chosen, leaving us 9-1= 8 choices)
Order of the non-matching cards doesn't matter, so we divide by 2!: 10*8/2! = 40.

Multiplying, we have 6*40=240 ways to combine a pair of the same value with a pair of different values.

Total number of ways to choose 2 pairs, each of which is of the same value:
We have 6 pairs to choose from, and the order of the pairs doesn't matter, so 6C2 = 15.

So total number of ways to get at least 1 pair of the same value = 240+15 = 255.

P(at least 1 pair of the same value) = 255/495 = 51/99 = 17/33.
Last edited by GMATGuruNY on Sun Oct 03, 2010 7:31 pm, edited 1 time in total.
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by tester_twelve » Sun Oct 03, 2010 7:29 pm
GMATGuruNY wrote:
tester_twelve wrote:
Ian Stewart wrote:
jason49 wrote: I think that C is correct, trying to figure out what tgmat calculated, because it obviously includes more cases than necessary.

Can somebody comment on this?
Nice find. In probability questions, when you see the words 'at least', as in this question, that often means that you would need to consider a few different cases if you did the problem directly. So this is often a clue that it will be faster to do the opposite problem: determine the probability that the event does not happen, then subtract from 1. That's exactly what sudhir did above, and that's exactly how I'd approach the problem as well.

That said, it is possible to do the problem directly, just as long as we break down the problem correctly. This solution is not quite right:
torontogmat.com wrote:It does not matter which card is turned first.

There is a 1/11 chance that the next card turned will match. If it doesn't,
there is a 2/10 chance that the following card will match. If it still doesn't,
there is a 3/9 chance that the final card will match.

1/11 + 1/5 + 1/3 = 103 / 165.
The probability, for example, that the last card matches one of the first three is not equal to 1/3, which is what the above solution assumes. It is only equal to 1/3 when we know that none of the first three cards match. We can adjust this solution to get the correct answer, however:

We get at least one pair IF:

[1+2 match] OR [(1+2 do not match)AND(1/2 matches 3)] OR [(1,2,3 do not match)AND(4 matches 1,2 or 3)]

The above solution missed the AND conditions. Solving:

1/11 + (10/11)(1/5) + [(10/11)(8/10)](1/3) = 15/165 + 30/165 + 40/165 = 85/165 = 17/33

Still, that's more difficult than the approach sudhir took above, which is much better.
Hi Ian/others,

Is there a way to solve this problem by first starting out with the total number of ways 4 cards can be drawn as the denominator (i.e. 12C4) and then determine what we're looking for in the numerator?
Yes, although it's not the approach that I would recommend.

Total number of ways to choose 12 cards = 12C4 = 495.

To get at least 1 pair of the same value, we need either to combine 1 pair of the same value with 1 pair of different values or to combine 2 pairs, each of which is of the same value.

Total number of ways to choose exactly 1 pair of the same value:
Number of ways to choose a pair of the same value = 6

Number of ways to choose a pair of different values from the 10 remaining cards:
1st card = 10 choices
2nd card = 8 choices (9 cards left, but we can't choose the mate of the card just chosen, leaving us 9-1= 8 choices)
Order of the non-matching cards doesn't matter, so we divide by 2!: 10*8/2! = 40.

Multiplying, we have 6*40=240 ways to combine a pair of the same value with a pair of different values.

Total number of ways to choose 2 pairs, each of which is of the same value:
We have 6 pairs to choose from, and the order of the pairs doesn't matter, so 6C2 = 15.

So total number of ways to get at least 1 pair of the same value = 240+15 = 255.

P(at least 1 pair of the same value) = 255/495 = 51/99 = 17/33.
Awesome. Thanks GMATGuruNY! And yes, while I have reviewed the other ways to go about solving this problem in this post and see that there are more strategic ways to solve this more quickly (which I will definitely try to employ on test day), it helps my probability polishing to other ways like this one.
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