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For those of you familiar with poker, how many different

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For those of you familiar with poker, how many different [#permalink] New post 29 Jul 2003, 15:48
For those of you familiar with poker, how many different 5-card hands will contain only one pair (XXWYZ, two of something and 3 other cards not matching the first two nor each other; e.g., AAKQ9 is a pair of aces).

How many 5-card hands will contain two-pairs (XXYYZ, two of something, two of something else, and a third card not matching either)?

Both will be relatively big numbers so leave them in factorial form.
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 [#permalink] New post 29 Jul 2003, 16:45
If i have understiid the problem correct then this is my try for time being on the first question:

1. Getting a pair formed of a particular type of Card (for eg: ACE) is 4C2 ways = 6 ways.

So there are 13 such different cards.

So, it will be 6*13 ways.

Now for choosing a card from the rest 48 cards, since 4 are gone above in form of paris. (only one pair allowed). So 48C1 = 48.

Now from select a card from rest 44 cards since 4 cards gone again as one type is chosen above so 44C1.

Now for the last card we have 40 cards left, going by the above logic. So 40C1.

Ans = 78*48*44*40 ways.

Is that fine...

IF yes, then is shall go ahead and solve the second one :wink:
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 [#permalink] New post 29 Jul 2003, 23:02
evensflow wrote:
If i have understiid the problem correct then this is my try for time being on the first question:

1. Getting a pair formed of a particular type of Card (for eg: ACE) is 4C2 ways = 6 ways.

So there are 13 such different cards.

So, it will be 6*13 ways.

Now for choosing a card from the rest 48 cards, since 4 are gone above in form of paris. (only one pair allowed). So 48C1 = 48.

Now from select a card from rest 44 cards since 4 cards gone again as one type is chosen above so 44C1.


Now for the last card we have 40 cards left, going by the above logic. So 40C1.

Ans = 78*48*44*40 ways.

Is that fine...

IF yes, then is shall go ahead and solve the second one :wink:


Nice try. Remember that a poker hand is a combination, not a permutation. (this can be a difficult problem, but if you truly understand how to count, it should not be too bad).
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Answer [#permalink] New post 30 Jul 2003, 07:17
I'll take a shot!!

1st card: 52 (can be any card in the deck)
2nd card: 3 (must complete the pair of the first card)
3rd card: 48 (can be any card in the deck that does not match the first two)
4th card: 44 (can be any card that does not match the first three)
5th card 40 (can be any card in the deck that does not match the first 4)

Finally, I believe the answer is (52*3*48*44*40)/5! to account for the fact that order does not matter. (Normally I would divide by 4! when two items are the same in a group of five, but in this instance the two cards in the pair are distinguishable, by suits)

Now let me ponder the second one...
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Answer [#permalink] New post 30 Jul 2003, 07:26
Ok, #2...

1st card: 52 (can be any card)
2nd card: 3 (must match first)
3rd card: 48 (can be any card but the first)
4th card: 3 (must match 3rd)
5th card: 44 (can be any other card)

SO... (52*3*48*3*44)/5! (again divided by 5! because all 5 members of the group are still distinguishable)
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Re: Answer [#permalink] New post 30 Jul 2003, 11:09
mciatto wrote:
I'll take a shot!!

1st card: 52 (can be any card in the deck)
2nd card: 3 (must complete the pair of the first card)
3rd card: 48 (can be any card in the deck that does not match the first two)
4th card: 44 (can be any card that does not match the first three)
5th card 40 (can be any card in the deck that does not match the first 4)

Finally, I believe the answer is (52*3*48*44*40)/5! to account for the fact that order does not matter. (Normally I would divide by 4! when two items are the same in a group of five, but in this instance the two cards in the pair are distinguishable, by suits)

Now let me ponder the second one...


You divide by 5! which means you are dividing by every possible permutation of a specific hand, say:

Ah Ad Ks 8h 4d.

However, your generating algorithm only generates hands in which the pair is in the first two slots. Hence, you are dividing by too much.

Try again....
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Answer [#permalink] New post 30 Jul 2003, 11:29
Well, my second choice is to divide by 5C2, as we have 10 different possible combinations of "placements" of the pair.
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Re: Answer [#permalink] New post 30 Jul 2003, 18:42
mciatto wrote:
Well, my second choice is to divide by 5C2, as we have 10 different possible combinations of "placements" of the pair.


Solution:

Method 1:
First, pick the value of the pairs from the 13 possible values. For whatever value we choose, we need to draw 2 of the 4 cards of that value. So we have: 13 * 4C2 = 13 * 6 so far. Next we need to pick 3 values out of the 12 remaining to decide what value each of the next 3 cards are. This is 12C3. Then for each of the values, we can pick from any of 4 suits, or 4*4*4.

Hence, the final result is 13 * 4C2 * 12C3 * 4^3 = 1,098,240


Method 2:First, pick one of any 52 cards. Whatever card you pick, there are 3 cards that match. However, since pulling the As first, then the Ac second is the same as pulling the Ac first and the As second, we need to adjust this by dividing by 2!. Hence, we have so far:

52 * 3 / 2! = 52 * 3 / 2.

For the next three cards, we have 48 * 44 * 40 ways to pick them. However, Ks Jc 10d, Jc 10d Ks, etc are equivalent, so we need to divide by 3! or 6.

Hence the final result is: (52 * 3 / 2) * (48 * 44 * 40) / 6 = 26 * 3 * 8 * 44 * 40 = 1,098,240.


Let's look at Mciatto's method:take any of 52 cards, then find a pair for it. Then take any of 48, then 44, then 40 cards. This is:

52 * 3 * 48 * 44 * 40.

Now he says, divide by 5!. This is wrong. The only time you need to divide by 5! is when you have generated all 5! possibilities of the same hand. In this case, however, we have only generated the situation where the paired cards are in the first two positions.

In order to generate ALL possible positions, we need to know how many ways the pair can be distributed in the 5-card hand. That is simply 5C2 or 10. Now we need to MULTIPLY our result by 5C2 to get the total number of permutation, then divide by 5!.

So we get (52 * 3 * 48 * 44 * 40) * 10 / 5! = 1,098,240.
Note how awkward it is to take a permutation approach to a combination problem.

Mciatto could also have done this.
take any of 52 cards, then find a pair for it. Then take any of 48, then 44, then 40 cards. This is:

52 * 3 * 48 * 44 * 40.

Now at this point, there are two ways the pairs can combine in slots one and two, and 3! ways the other 3 can combine in slots 3, 4, and 5. Hence, we just adjust for those and we divide first by 2!, then by 3!. New result:

52 * 3 * 48 * 44 * 40 / (2 * 6) = 1,098,240.

This result is exactly equivalent to method 2.


I like this problem because if you really understand it, you really understand the nuances of counting and double-counting and the multiplication rule. My advise is to take this problem apart, try to solve it yourself, then try to completely understand all three methods. Note where your assumptions regarding counting were right, and note where what you thought was right is wrong. If you invest the time in undertstanding COMPLETELY how to count this one problem, you will fully understand the bulk of what combinatorics is all about.

Anybody want to try #2 now?
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 [#permalink] New post 01 Aug 2003, 10:45
No takers? I'm dissapointed. You see, the way to learn all the nuances to combinatorics is to take a tough problem, try to solve it using everything you know about combinations and permutations, then find out where your assumption went wrongs, and internalize it.

You can spent 8 zillion years solving pure combination/permutation problems, but until you can solve just one problem like this, you will never REALLY understand what is going on.

Solution:

First you need to pick the two values that you will use for the two pair out of the 13 possible values. That is 13C2. Then you need to pick 2 out of the possible 4 suits for each of the two values. Finally you need to pick a card from the 44 that remain that don't match either value.

13C2 * 4C2 * 4C2 * 44 = 78 * 6 * 6 * 44 = 123552.

Simple as that. No complicated formulas. Just logical thinking.

There are at least 2 other ways to approach this problem, similar to the alternate solutions for part 1.

P.S. The "hot" thread here is the constant whining about how hard the quant is on the REAL test compared to the stuff in the OG, Kaplan, PR, etc. Yet, nobody is making the effort to solve the "tough" problems i put here. More satisfying to blame ETS and moan, i suppose. No sympathy from me.

Axioms:
1) you cannot buy a score
2) ETS is not testing your math ability. It is testing your ability to solve an unlimited range of problems with a limited number of tools.
3) Of course, ETS purposely gives your problems unlike anything you've seen on OG or any prep course. But the math you need to know is never more than basic.
4) Solving ONE tough problem in a particular area and FULLY UNDERSTANDING becasue you spent some serious time taking it apart and putting it back together is MUCH more educational than solving 100 stupid Kaplan drills.

You want to understand probabiltiy? Do this. Spend a few hours calcuating the probabiltiy of getting every possible poker hand: Royal flush, straight flush, full house, straight, flush, 3of akind, two pair. pair, nothing. Once you understand counting, this is a piece of cake. And if you can do this, then you REALLY understand counting.

Next calculate the probability of winning a straight $10 bet on the pass line in a game a craps. This is not difficult, just time consuming and you need to be systematic. However, just working (and i mean working) through these two problems will give you an incredible understanding in combinatorics and probability - more than most will learn in a high school class. Once you understand this, the stupid Kaplan and PR prob problem will be cake. More importantlly, you will be prepared to tackle a real GMAT problem rated "very difficult".

Or you can just ignore the tough problems, skip the hard work, and whine.
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Probability for Akamai [#permalink] New post 19 Aug 2003, 17:52
AkamaiBrah wrote:
No takers? I'm dissapointed. You see, the way to learn all the nuances to combinatorics is to take a tough problem, try to solve it using everything you know about combinations and permutations, then find out where your assumption went wrongs, and internalize it.

You can spent 8 zillion years solving pure combination/permutation problems, but until you can solve just one problem like this, you will never REALLY understand what is going on.

Solution:

First you need to pick the two values that you will use for the two pair out of the 13 possible values. That is 13C2. Then you need to pick 2 out of the possible 4 suits for each of the two values. Finally you need to pick a card from the 44 that remain that don't match either value.

13C2 * 4C2 * 4C2 * 44 = 78 * 6 * 6 * 44 = 123552.

Simple as that. No complicated formulas. Just logical thinking.

There are at least 2 other ways to approach this problem, similar to the alternate solutions for part 1.




hello

i tried the other way, but i think i need some help

i got 52*3/2 * 48*3/2 * 44

i think that pair 1 and pair 2 can be arranged in 2! ways...if i take into account this factor, then i get the right answer..

am i right or did i miss something?

thanks a lot for your explanation
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Re: Probability for Akamai [#permalink] New post 19 Aug 2003, 23:13
praetorian123 wrote:
AkamaiBrah wrote:
No takers? I'm dissapointed. You see, the way to learn all the nuances to combinatorics is to take a tough problem, try to solve it using everything you know about combinations and permutations, then find out where your assumption went wrongs, and internalize it.

You can spent 8 zillion years solving pure combination/permutation problems, but until you can solve just one problem like this, you will never REALLY understand what is going on.

Solution:

First you need to pick the two values that you will use for the two pair out of the 13 possible values. That is 13C2. Then you need to pick 2 out of the possible 4 suits for each of the two values. Finally you need to pick a card from the 44 that remain that don't match either value.

13C2 * 4C2 * 4C2 * 44 = 78 * 6 * 6 * 44 = 123552.

Simple as that. No complicated formulas. Just logical thinking.

There are at least 2 other ways to approach this problem, similar to the alternate solutions for part 1.




hello

i tried the other way, but i think i need some help

i got 52*3/2 * 48*3/2 * 44

i think that pair 1 and pair 2 can be arranged in 2! ways...if i take into account this factor, then i get the right answer..

am i right or did i miss something?

thanks a lot for your explanation
praetorian


You are right. KKQQ and QQKK are the same combination, but you count them twice.
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DS Question? [#permalink] New post 21 Aug 2003, 06:49
On a certain day it took Bill three times as long to drive from home to work as it took Sue to drive from home to work. How many kilometers did Bill drive from home to work?
(1) Sue drove 10 kilometers from home to work, and the ratio of distance driven from home to work time to drive from home to work was the same for Bill and Sue that day.
(2) The ratio of distance driven from home to work time to drive from home to work for Sue that day was 64 kilometers per hour.

Icould solve the problem but my main idea is to get a little clarification on the wording i.e. "ratio of distance driven from home to work time to drive from home to work was the same for Bill and Sue that day" how to put this statement into a ratio and whate exactly is the literal meaning of this statement[/quote]
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PS THREAD 1 [#permalink] New post 25 Aug 2003, 15:41
On a Saturday night, each of the rooms at a certain motel was rented for either $40 or $60. If 10 of the rooms that were rented for $60 had instead been rented for $40, then the total rent the motel charged for that night would have been reduced by 25 percent. What was the total rent the motel actually charged for that night ?
(A) $600
(B) $800
(C) $1,000
(D) $1,600
(E) $2,400
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PS THREAD 2 [#permalink] New post 25 Aug 2003, 16:24
Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of this mixture is X ?
(A) 10%
(B) 33.33 %
(C) 40%
(D) 50%
(E) 66.66 %
PL. EXPLAIN
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Re: PS THREAD 2 [#permalink] New post 25 Aug 2003, 17:13
aps_can wrote:
Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of this mixture is X ?
(A) 10%
(B) 33.33 %
(C) 40%
(D) 50%
(E) 66.66 %
PL. EXPLAIN


Please post your questions in a new thread .. PLEASE !!

Now for your question ..
Say new mixture contains x % from X and y% from Y
hence Ryegrass = 0.4x + 0.25y which is 30% of (x+y)
==> 2x = y
that means for every 'y' , we need 2x ..
so % of x must be 1/3 x 100 = 33.33%

as a sanity check , say I select 100gms of Y ,
which cintains 25 gms of Rye grass..
Now i need 200 gms of X
which contains 80gms of Ryegrass and 120gms of bluegrass
Total Ryegrass = 90
Total mixture = 300
so Ryegrass % = 90/300 x 100 = 30%
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Re: PS THREAD 2   [#permalink] 25 Aug 2003, 17:13
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