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A card game called “high-low” divides a deck of 52 playing c

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A card game called “high-low” divides a deck of 52 playing c  [#permalink]

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Updated on: 10 Jun 2014, 01:24
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A card game called “high-low” divides a deck of 52 playing cards into 2 types, “high” cards and “low” cards. There are an equal number of “high” cards and “low” cards in the deck and “high” cards are worth 2 points, while “low” cards are worth 1 point. If you draw cards one at a time, how many ways can you draw “high” and “low” cards to earn 5 points if you must draw exactly 3 “low” cards?

A. 1
B. 2
C. 3
D. 4
E. 5

Originally posted by ravih on 09 Jun 2014, 19:51.
Last edited by Bunuel on 10 Jun 2014, 01:24, edited 1 time in total.
Edited the question.
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Re: A card game called “high-low”  [#permalink]

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09 Jun 2014, 21:31
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Great question Ravih. This is a permutations problem (order matters) with repeating elements. Given that "low" cards are worth 1 pt and "high cards" 2 pts, and you must draw 3 low cards, we know that you must also draw 1 high card. The formula for permutations problems with repeating elements is "N!/A!B!..." where N represents the number of elements in the group and A, B, etc. represent the number of times that repeating elements are repeated. Here there are 4 elements and the "low" card is repeated 3 times. As a result, the formula is:

4!/3! which represents (4*3*2*1)/(3*2*1) which simplifies to just 4, giving you answer D.

Another way to think about it that you must have 3 low cards and one high card, and the question is asking you about the different possible orders. The 4 orders are:

H-L-L-L, L-H-L-L, L-L-H-L, and finally L-L-L-H, giving you 4 options.

I hope this helps!!!
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Re: A card game called “high-low”  [#permalink]

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10 Jun 2014, 00:04
Thank you Brandon for the explanation.

However, I still have some confusion. I believe that this is a combination problem not a permutation one as the question basically asks us to choose cards (1 from the 26 High cards & 3 from the 26 Low cards). The arrangement of the cards is not moot. Therefore the answer should be:
26C1 x 26C3

Following is the solution from Kaplan which I somehow did not find very convincing. Although they have confirmed that this is a combination problem, they have applied the formula to only the 4 cards and not the selection from the entire batch of 52 cards:

"First, we need to figure out how many cards of each type we will need in order to earn 5 points. The "low" cards are worth 1 point. Since the question says we must draw 3 of these "low" cards, we get 3 points from "low" cards. This leaves us with 5 – 3 = 2 points left that we need to earn with the "high" cards. Since "high" cards are worth 2 points each, if we get one "high" card, we will reach our goal of 5 points total. Hence it takes us four cards to get 5 points: 3 "low" and 1 "high."
Our task is now simply to discover how many ways we can draw these 4 cards. Let’s draw out the possible ways: (L L L H), (L L H L), (L H L L), (H L L L) for a total of 4 different ways to get 3 "low" and 1 "high." Answer Choice (D) is correct.
Alternatively, we could have used the combination formula, which is the number of different ways of choosing k objects from n different objects. If nCk is the number of different subgroups of k objects that can be selected from a group of n different objects, then . Remember, in this formula,n ≥ k.
Now let’s plug numbers into the formula so that n is greater than or equal to k. We know we need 4 cards, so n = 4. Also, k is always less than or equal to n, and in this case we know we need 3 of the "low" cards, so k = 3. The number of possibilities is
This method also brings us to Answer Choice (D). The combination formula is the concrete math underlying this problem, but the key step here is realizing that it is unnecessary; it is much faster and easier just to count the possibilities."
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Re: A card game called “high-low”  [#permalink]

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11 Jun 2014, 01:22
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ravih wrote:
Thank you Brandon for the explanation.

However, I still have some confusion. I believe that this is a combination problem not a permutation one as the question basically asks us to choose cards (1 from the 26 High cards & 3 from the 26 Low cards). The arrangement of the cards is not moot. Therefore the answer should be:
26C1 x 26C3

Following is the solution from Kaplan which I somehow did not find very convincing. Although they have confirmed that this is a combination problem, they have applied the formula to only the 4 cards and not the selection from the entire batch of 52 cards:

"First, we need to figure out how many cards of each type we will need in order to earn 5 points. The "low" cards are worth 1 point. Since the question says we must draw 3 of these "low" cards, we get 3 points from "low" cards. This leaves us with 5 – 3 = 2 points left that we need to earn with the "high" cards. Since "high" cards are worth 2 points each, if we get one "high" card, we will reach our goal of 5 points total. Hence it takes us four cards to get 5 points: 3 "low" and 1 "high."
Our task is now simply to discover how many ways we can draw these 4 cards. Let’s draw out the possible ways: (L L L H), (L L H L), (L H L L), (H L L L) for a total of 4 different ways to get 3 "low" and 1 "high." Answer Choice (D) is correct.
Alternatively, we could have used the combination formula, which is the number of different ways of choosing k objects from n different objects. If nCk is the number of different subgroups of k objects that can be selected from a group of n different objects, then . Remember, in this formula,n ≥ k.
Now let’s plug numbers into the formula so that n is greater than or equal to k. We know we need 4 cards, so n = 4. Also, k is always less than or equal to n, and in this case we know we need 3 of the "low" cards, so k = 3. The number of possibilities is
This method also brings us to Answer Choice (D). The combination formula is the concrete math underlying this problem, but the key step here is realizing that it is unnecessary; it is much faster and easier just to count the possibilities."

Following up on Brandon's explanation, let me point out a few things:

"A card game called “high-low” divides a deck of 52 playing cards into 2 types, “high” cards and “low” cards."

Note that this is a different card game. It divides a deck of 52 cards into two types - high and low. Each card is not unique. Do not think of them as the deck of 52 unique cards of different colors and numbered 1 to 13 with which we play! This game has 52 cards divided into two groups - high and low. If you pick the high card, you get 2 point. If you pick the low card, you get 1 point.

Now we know we need to pick 4 cards and three of them have to be low and 1 has to be high. We are given in the question that the cards are picked one after the other (not together). In how many ways can we pick 3 low and 1 high cards?
You can do it in 4 ways: HLLL, LHLL, LLHL, LLLH

This question is quite similar to coin toss questions - in how many ways can we get 1 Head and 3 Tails on 4 coin tosses?

So at the end of it all, we can say that it is actually a permutation question! 4 cards need to be arranged such that 3 cards are the same: 4!/3! = 4

Though something can be said about the language of the question. 52 playing cards makes one think that we are talking about the regular playing cards. But, the solution given by Brandon is correct and it is the intent of the question as is apparent from Kaplan's solution too.
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Re: A card game called “high-low” divides a deck of 52 playing c  [#permalink]

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16 Oct 2014, 13:08
ravih wrote:
A card game called “high-low” divides a deck of 52 playing cards into 2 types, “high” cards and “low” cards. There are an equal number of “high” cards and “low” cards in the deck and “high” cards are worth 2 points, while “low” cards are worth 1 point. If you draw cards one at a time, how many ways can you draw “high” and “low” cards to earn 5 points if you must draw exactly 3 “low” cards?

A. 1
B. 2
C. 3
D. 4
E. 5

To get a 5, you need one high and three lows (you could have had 2 highs and one low, but the constraint is that you must have three low cards)
HLLL = 4!3! =4

4! is the number of ways you can arrange these four spaces. divide by 3! because you you repeat three low cards. 30 second solution

+1 if I helped!
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Re: A card game called “high-low” divides a deck of 52 playing c  [#permalink]

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18 Oct 2014, 00:44
ravih wrote:
A card game called “high-low” divides a deck of 52 playing cards into 2 types, “high” cards and “low” cards. There are an equal number of “high” cards and “low” cards in the deck and “high” cards are worth 2 points, while “low” cards are worth 1 point. If you draw cards one at a time, how many ways can you draw “high” and “low” cards to earn 5 points if you must draw exactly 3 “low” cards?

A. 1
B. 2
C. 3
D. 4
E. 5

HLLL = 4!/3! = 4 ways
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Re: A card game called “high-low” divides a deck of 52 playing c  [#permalink]

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13 Nov 2016, 03:36
ravih wrote:
A card game called “high-low” divides a deck of 52 playing cards into 2 types, “high” cards and “low” cards. There are an equal number of “high” cards and “low” cards in the deck and “high” cards are worth 2 points, while “low” cards are worth 1 point. If you draw cards one at a time, how many ways can you draw “high” and “low” cards to earn 5 points if you must draw exactly 3 “low” cards?

A. 1
B. 2
C. 3
D. 4
E. 5

Why don't we consider the option LLL-L-L ? This adds +1 to the LLL-H ways (4) and in the end we have 5 ways
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Re: A card game called “high-low” divides a deck of 52 playing c  [#permalink]

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14 Nov 2016, 05:03
chismooo wrote:
ravih wrote:
A card game called “high-low” divides a deck of 52 playing cards into 2 types, “high” cards and “low” cards. There are an equal number of “high” cards and “low” cards in the deck and “high” cards are worth 2 points, while “low” cards are worth 1 point. If you draw cards one at a time, how many ways can you draw “high” and “low” cards to earn 5 points if you must draw exactly 3 “low” cards?

A. 1
B. 2
C. 3
D. 4
E. 5

Why don't we consider the option LLL-L-L ? This adds +1 to the LLL-H ways (4) and in the end we have 5 ways

You need to draw EXACTLY 3 low cards. So you cannot draw more than 3 low cards.
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Re: A card game called “high-low” divides a deck of 52 playing c  [#permalink]

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19 Jun 2017, 07:12
ravih wrote:
A card game called “high-low” divides a deck of 52 playing cards into 2 types, “high” cards and “low” cards. There are an equal number of “high” cards and “low” cards in the deck and “high” cards are worth 2 points, while “low” cards are worth 1 point. If you draw cards one at a time, how many ways can you draw “high” and “low” cards to earn 5 points if you must draw exactly 3 “low” cards?

A. 1
B. 2
C. 3
D. 4
E. 5

I would say that the question is wrong. It doesn't specify the point that all the high cards are similar and all the low cards are similar. Only in that case would the answer be 4. Otherwise, the correct answer is not mentioned in the options. Please don't make ambiguous questions, as it scares people.
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Re: A card game called “high-low” divides a deck of 52 playing c  [#permalink]

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12 Feb 2018, 13:05
Hi All,

Some of the information in this prompt is ultimately not a factor in the solution (the fact that there are 26 'high' cards and 26 'low' cards is not a factor - it just establishes that there are enough cards to score "5 points").

We're told that high cards are worth 2 points and low cards are worth 1 point. Then we're told that we score 5 points when drawing exactly 3 low cards. Since those 3 low cards are worth a total of 3(1) = 3 points, the remaining 2 points MUST come from a high card.

Thus, we have 3 low cards and 1 high card. Drawing cards one at a time, there are only a certain number of ways to score 5 points under these conditions:

HLLL
LHLL
LLHL
LLLH

4 ways.

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Re: A card game called “high-low” divides a deck of 52 playing c  [#permalink]

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10 Aug 2018, 18:44
ravih wrote:
A card game called “high-low” divides a deck of 52 playing cards into 2 types, “high” cards and “low” cards. There are an equal number of “high” cards and “low” cards in the deck and “high” cards are worth 2 points, while “low” cards are worth 1 point. If you draw cards one at a time, how many ways can you draw “high” and “low” cards to earn 5 points if you must draw exactly 3 “low” cards?

A. 1
B. 2
C. 3
D. 4
E. 5

To get 5 points and drawing 3 “low” cards, you must draw exactly 1 “high” card also. One of the ways this can be done is LLLH. However, there are 4!/3! = 4 ways to arrange 3 L’s and 1 H. Therefore, there are 4 ways this can be done.

Alternate Solution:

We can list every possible ways to obtain 5 points where exactly 3 low cards are drawn:

LLLH, LLHL, LHLL, HLLL

We see that there are 4 ways.

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Re: A card game called “high-low” divides a deck of 52 playing c   [#permalink] 10 Aug 2018, 18:44
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