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A deck of cards contains 6 cards numbered from 1 to 6. If

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A deck of cards contains 6 cards numbered from 1 to 6. If [#permalink]

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08 Nov 2011, 02:26
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A deck of cards contains 6 cards numbered from 1 to 6. If three cards are randomly selected from the deck, what is the probability that the numbers on the cards are drawn in ascending order (example: first draw: 3, second draw: 4, third draw: 5)

[Reveal] Spoiler:
I don´t get it and would appreciat th help.
My approach was: Denominator: (6*5*4*/3!) = 20
Numerator: I just know, that there are four groups of ascending numbers (1,2,3)(2,3,4)(3,4,5)(4,5,6). But I am not sure yet, how to express this.

Any ideas?
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Re: Numbered cards...drawing in ascending order [#permalink]

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08 Nov 2011, 02:54
This is actually not that hard... Your approach is just fine, but you made a mistake when calculating the total number of possible card combinations. It is:

C = 6!/(6-3)! = 6!/3! = 4*5*6 = 120

Hence, the probability is P = 4/120 = 1/30.
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Re: Numbered cards...drawing in ascending order [#permalink]

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08 Nov 2011, 03:00
Sepideh2612 wrote:
A deck of cards contains 6 cards numbered from 1 to 6. If three cards are randomly selected from the deck, what is the probability that the numbers on the cards are drawn in ascending order (example: first draw: 3, second draw: 4, third draw: 5)

I don´t get it and would appreciat th help.
My approach was: Denominator: (6*5*4*/3!) = 20
Numerator: I just know, that there are four groups of ascending numbers (1,2,3)(2,3,4)(3,4,5)(4,5,6). But I am not sure yet, how to express this.

Any ideas?

What are the options?
I got 1/5 as the answer
denominator : no. of ways of selecting 3 cards = 6C3
numerator : selecting the cards in ascending order = {1,2,3},(2,3,4},{3,4,5},{4,5,6}= 4
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Re: Numbered cards...drawing in ascending order [#permalink]

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08 Nov 2011, 03:10
I'd also be interested in the source of the questions you posted here. Thanks.
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Re: Numbered cards...drawing in ascending order [#permalink]

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08 Nov 2011, 03:26
dmnk wrote:
This is actually not that hard... Your approach is just fine, but you made a mistake when calculating the total number of possible card combinations. It is:

C = 6!/(6-3)! = 6!/3! = 4*5*6 = 120

Hence, the probability is P = 4/120 = 1/30.

Hi,
How do you get the denominator as 6p3 when ur selecting 3 out of 6 then it should be 6C3 .
and the numerator : 4

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Re: Numbered cards...drawing in ascending order [#permalink]

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08 Nov 2011, 03:50
kalrac wrote:
Hi,
How do you get the denominator as 6p3 when ur selecting 3 out of 6 then it should be 6C3 .
and the numerator : 4

Sorry, I'm not familiar with the expressions "6P3" and "6C3" you used. Do they refer to permutations as opposed to combinations?

The order of the cards must be considered here, e.g. both {1,2,3} and {2,1,3} are possible outcomes, but only the first is in ascending order.
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Re: Numbered cards...drawing in ascending order [#permalink]

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08 Nov 2011, 04:17
dmnk wrote:
kalrac wrote:
Hi,
How do you get the denominator as 6p3 when ur selecting 3 out of 6 then it should be 6C3 .
and the numerator : 4

Sorry, I'm not familiar with the expressions "6P3" and "6C3" you used. Do they refer to permutations as opposed to combinations?

The order of the cards must be considered here, e.g. both {1,2,3} and {2,1,3} are possible outcomes, but only the first is in ascending order.

I was referring to permutations and combinations i.e arrangements and selections.

the qn asks to select 3 out of 6 and not arrange in 3 from 6.
So the tot no. of selections in random would be 20 and not 120.
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Re: Numbered cards...drawing in ascending order [#permalink]

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08 Nov 2011, 04:42
kalrac wrote:
dmnk wrote:
kalrac wrote:
Hi,
How do you get the denominator as 6p3 when ur selecting 3 out of 6 then it should be 6C3 .
and the numerator : 4

Sorry, I'm not familiar with the expressions "6P3" and "6C3" you used. Do they refer to permutations as opposed to combinations?

The order of the cards must be considered here, e.g. both {1,2,3} and {2,1,3} are possible outcomes, but only the first is in ascending order.

I was referring to permutations and combinations i.e arrangements and selections.

the qn asks to select 3 out of 6 and not arrange in 3 from 6.
So the tot no. of selections in random would be 20 and not 120.

The question is not well-written... Still, as I see it, by selecting three out of six cards you automatically arrange them, because you have a first draw, second draw and third draw. The first card drawn will remain in the first spot, so to speak. If you select card 1 in draw 1, card 2 in draw 2 and card 3 in draw 3, this gives {1,2,3}. It does not give {2,3,1}, {3,1,2} etc.

I'm interested in the source of the question. I'm pretty sure it's no official GMAC question and the MGMAT questions I know seem better, too.
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Re: Numbered cards...drawing in ascending order [#permalink]

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08 Nov 2011, 07:41
Uhmm .. IMO this Question is a little ambiguously written. for the ascending part... I believe that it doesnt mean they have to be consecutive numbers, such as 3,4 and 5, also 3,4,6 complies with the conditions, so ..

This is my approach to this problem:

1. Denominator: 6C3 = 6x5x4 = 120 = total possible way of drawing the 3 cards.
2. numerator: I was not really sure how to approach this using combinations or permutations so I quickly wrote all possibilities (it was not that bad):

if I draw a 1 in the first draw then my possibilities are:
123 134 145 156
124 135 146
125 136
126

then if I drew a 2 on my first round :

234 245 256
235 246
236

for 3:

345 356
346

and finally for 4:

456

if I drew a 5 or a 6 on the first draw it is impossible to get ascending numbers in the following draws.

That said i just counted the "positive" cases for the numerator: 20

and, divided that by the denominator and that was it:

successful cases / total cases = 20/120 = 1/6.

It would help a lot to have options here... please let us know the source.

Thanks!

PS: If this helped you please Kudo'me....
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Re: Numbered cards...drawing in ascending order [#permalink]

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08 Nov 2011, 10:06
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Sepideh2612 wrote:
A deck of cards contains 6 cards numbered from 1 to 6. If three cards are randomly selected from the deck, what is the probability that the numbers on the cards are drawn in ascending order (example: first draw: 3, second draw: 4, third draw: 5)

I don´t get it and would appreciat th help.
My approach was: Denominator: (6*5*4*/3!) = 20
Numerator: I just know, that there are four groups of ascending numbers (1,2,3)(2,3,4)(3,4,5)(4,5,6). But I am not sure yet, how to express this.

Any ideas?

There is certainly some ambiguity here but the following seems to be the intention of the question:
A deck of cards contains 6 cards numbered from 1 to 6. If three cards are randomly selected one-by-one from the deck, what is the probability that the numbers on the cards drawn are consecutive and in ascending order?

To get 3 consecutive ascending order numbers, I should draw 1/2/3/4 on the first draw. Probability = 4/6
Out of the remaining 5, there is only one choice for the next card (If I drew 2 on the first go, I must draw 3 now). Probability = 1/5
Out of the remaining 4, there is only one choice for the lats card (If I drew 3 on the second go, I must draw 4 now). Probability = 1/4

Probability of drawing 3 consecutive numbers in ascending order = (4/6)*(1/5)*(1/4) = 1/30
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Re: Numbered cards...drawing in ascending order [#permalink]

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08 Nov 2011, 10:41
Hi guys,

thanks for the help. There was actually the problem: I didn´t consider the order, that´s why I divided by 3! to get rid of the permutations.
I understand it now completely thanks to you.

I have the question from a CD I got from a GMAT institute in Germany. Most of the questions are copied from some GMAT books, I guess, because most of them were already discusse in most fores. Unfortunately, the ones, I posted, were not.
Maybe they are leading to misnderstanding. I didn´t want to confuse anybody, sorry!

Thanks a lot anyway!!!
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Re: Numbered cards...drawing in ascending order [#permalink]

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13 Dec 2011, 07:31
Could you plz tell me why $$\frac{1}{5}$$ is wrong?
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Re: Numbered cards...drawing in ascending order [#permalink]

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13 Dec 2011, 12:47
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Let me throw my 2 cents in... Like Karishma said, the question is ambiguous. There are two possible cases here, depending on how you pick.

First lets rephrase: 6 cards, 3 picks. There are four ascending choices {1,2,3}, {2,3,4}, {3,4,5} and {4,5,6}

Now, lets analyze how we pick

1. Random pick of three: This says that we are applying combination logic here. Therefore P = 4 / 6C3 = 1/5

2. Pick-by-pick of three: This screams permutation. Therefore
P = 4 / 6P3 = 1/30

Paying attention to how picks are done in the question is critical. Again, like Karishma said, this question is ambiguous.
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Re: Numbered cards...drawing in ascending order [#permalink]

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13 Dec 2011, 22:07
here is how I approached it
denominator : 6*5*4 = 120
Numerator. i made the foll grid to arrive at 20 as numerator

probability - 1/6... what is wrong with this

If 1 and 2 are the first cards, the third card drawn is 3 or 4 or 5 or 6 , therefore 4 ways
Similiarly if 1 and 3 are first 2 cards drawn then then 3rd card can be drawn in 3 ways and so on
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Re: Numbered cards...drawing in ascending order [#permalink]

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15 Dec 2011, 06:10
Sepideh2612 wrote:
[b]A deck of cards contains 6 cards numbered from 1 to 6. If three cards are randomly selected from the deck, what is the probability that the numbers on the cards are drawn in ascending order

The wording of the question is pretty bad, unfortunately. First the question should make clear that the selections are made *without replacement*, and even more importantly, the word 'ascending' has nothing whatsoever to do with 'consecutive' integers. 'Ascending' only means 'increasing'; the sequence 3, 7, 22 is in 'ascending order', for example. As Karishma clarified above, it seems the question intends to ask for the probability of selecting, without replacement, three *consecutive* integers in increasing order, but the question needs to make that clear. Real GMAT questions would never be unclear about this point, and if you see the word 'ascending' on the GMAT, you should only take that to mean 'increasing' - it doesn't mean 'consecutive'.

Danielphonics above correctly found the probability of picking three numbers in increasing order, consecutive or not. I think that's an interesting question, and there is a simpler approach. I'll change the wording and numbers completely for clarity. Say you're asked:

A computer randomly picks three distinct integers from between 1 and 1000 inclusive. If the first number picked is x, the second number picked is y, and the third number picked is z, what is the probability that x < y < z ?

So the question is asking the probability that the selections are made in increasing order. If you pick any three different numbers (from any set at all), you could select those 3 numbers in 3! = 6 different orders. Only 1 of those orders is increasing, so the probability you made the selection in increasing order must be 1/6. Notice that it makes no difference here whether we pick our numbers from a set of one thousand integers, or from a set of six cards - the answer will always be the same.
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Re: Numbered cards...drawing in ascending order [#permalink]

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03 Apr 2013, 03:44
shahideh wrote:
Could you plz tell me why $$\frac{1}{5}$$ is wrong?

This is the right answer when cards are drawn in ascending order, not necessarily consecutive numbers in ascending order.
probability that the numbers on the cards drawn are consecutive numbers in ascending order = 1/30
probability that the numbers on the cards drawn are in ascending order = 1/5

Let me give proof.
Sepideh2612 wrote:
A deck of cards contains 6 cards numbered from 1 to 6. If three cards are randomly selected from the deck, what is the probability that the numbers on the cards are drawn in ascending order (example: first draw: 3, second draw: 4, third draw: 5)

Any ideas?

From question, original answer presented and example given above we can infer that the correct question should be - "A deck of cards contains 6 cards numbered from 1 to 6. If three cards are randomly selected one by one from the deck, what is the probability that the numbers on the cards drawn are consecutive numbers in ascending order."

Proof 1:
Since the question imposes the restriction that the numbers on the cards drawn should be consecutive in ascending order, so we have -
1. number on the first card drawn is 1, 2, 3, or 4. Probability of this event E1 = 4/6
2. now we have 5 cards left and since we have already chosen our first card, so there is only one choice left to choose next card, a card with next consecutive number out of 5 total cards. Probability of this event E2 = 1/5.
3. same logic holds for 3rd card as in case 2. Probability of this event E3 = 1/4.
So, Probability of occurring event 1 AND event 2 AND event 3 = (4/6) * (1/5) * (1/4) = 1/30

If this question was only interested in the ascending order of the numbers on the cards selected, then size of sample space S, |S| = 6P3 = 6!/3! = 6*5*4 = 120. Now we have -
1. number on the first card drawn is 1 to 4 inclusive. total no. of ways = 4.
2. number on the second card drawn depends on the number on first card drawn. i.e. if number in first card is 1 then 2nd card should be (2 or 3 or 4 or 5) and if number in first card is 2 then we have 3, 4, 5 left to draw. So total number of ways = 4 + 3 + 2 + 1 = 10.
3. similar logic as in 2 holds for 3rd card also(one of 3, 4, 5, 6). So total number of ways of choosing 3rd card = 4 + 3 + 2 + 1 = 10.
So, the probability that the numbers on the cards drawn are in ascending order = (4 + 10 + 10)/120 = 1/5.

Proof 2:
Sorry, I haven't published my theorem yet.
Re: Numbered cards...drawing in ascending order   [#permalink] 03 Apr 2013, 03:44
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