December 14, 2018 December 14, 2018 09:00 AM PST 10:00 AM PST 10 Questions will be posted on the forum and we will post a reply in this Topic with a link to each question. There are prizes for the winners. December 14, 2018 December 14, 2018 10:00 PM PST 11:00 PM PST Carolyn and Brett  nicely explained what is the typical day of a UCLA student. I am posting below recording of the webinar for those who could't attend this session.
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 15 May 2014
Posts: 9

A card game called “highlow” divides a deck of 52 playing c
[#permalink]
Show Tags
Updated on: 10 Jun 2014, 00:24
Question Stats:
56% (01:47) correct 44% (02:03) wrong based on 259 sessions
HideShow timer Statistics
A card game called “highlow” 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
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by ravih on 09 Jun 2014, 18:51.
Last edited by Bunuel on 10 Jun 2014, 00:24, edited 1 time in total.
Edited the question.



Veritas Prep GMAT Instructor
Joined: 23 Oct 2013
Posts: 144

Re: A card game called “highlow”
[#permalink]
Show Tags
09 Jun 2014, 20:31
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: HLLL, LHLL, LLHL, and finally LLLH, giving you 4 options. I hope this helps!!!
_________________
Brandon Veritas Prep  GMAT Instructor
If you found this post helpful, please give me kudos!!!
Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.
Veritas Prep Reviews



Intern
Joined: 15 May 2014
Posts: 9

Re: A card game called “highlow”
[#permalink]
Show Tags
09 Jun 2014, 23:04
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: 26 C1 x 26 C3 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."



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8677
Location: Pune, India

Re: A card game called “highlow”
[#permalink]
Show Tags
11 Jun 2014, 00:22
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: 26 C1 x 26 C3 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 “highlow” 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.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Joined: 30 Mar 2013
Posts: 109

Re: A card game called “highlow” divides a deck of 52 playing c
[#permalink]
Show Tags
16 Oct 2014, 12:08
ravih wrote: A card game called “highlow” 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!



Manager
Joined: 22 Jan 2014
Posts: 175
WE: Project Management (Computer Hardware)

Re: A card game called “highlow” divides a deck of 52 playing c
[#permalink]
Show Tags
17 Oct 2014, 23:44
ravih wrote: A card game called “highlow” 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
_________________
Illegitimi non carborundum.



Current Student
Joined: 08 Aug 2016
Posts: 36

Re: A card game called “highlow” divides a deck of 52 playing c
[#permalink]
Show Tags
13 Nov 2016, 02:36
ravih wrote: A card game called “highlow” 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 LLLLL ? This adds +1 to the LLLH ways (4) and in the end we have 5 ways



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8677
Location: Pune, India

Re: A card game called “highlow” divides a deck of 52 playing c
[#permalink]
Show Tags
14 Nov 2016, 04:03
chismooo wrote: ravih wrote: A card game called “highlow” 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 LLLLL ? This adds +1 to the LLLH 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.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Senior Manager
Joined: 03 Apr 2013
Posts: 274
Location: India
Concentration: Marketing, Finance
GPA: 3

Re: A card game called “highlow” divides a deck of 52 playing c
[#permalink]
Show Tags
19 Jun 2017, 06:12
ravih wrote: A card game called “highlow” 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.
_________________
Spread some love..Like = +1 Kudos



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13081
Location: United States (CA)

Re: A card game called “highlow” divides a deck of 52 playing c
[#permalink]
Show Tags
12 Feb 2018, 12:05
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. Final Answer: GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 4294
Location: United States (CA)

Re: A card game called “highlow” divides a deck of 52 playing c
[#permalink]
Show Tags
10 Aug 2018, 17:44
ravih wrote: A card game called “highlow” 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. Answer: D
_________________
Scott WoodburyStewart
Founder and CEO
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions




Re: A card game called “highlow” divides a deck of 52 playing c &nbs
[#permalink]
10 Aug 2018, 17:44






