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A card game called “high-low� divides a deck of 52 playi

Expert replies
by BTGmoderatorDC » Mon Jul 30, 2018 5:25 pm

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

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
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Source: — Problem Solving |

by Jay@ManhattanReview » Mon Jul 30, 2018 10:26 pm
BTGmoderatorDC 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
Complying with the condition that exactly 3 "low" cards must be drawn to earn 5 points, we must draw only one "High" card (each High card values 2 points). Thus, only four cards must be drawn: 1 High and 3 Low. Since the order is not important, the High card can be drawn in any of the four drawings.

Possible four ways: HLLL, LHLL, LLHL, and LLLH

The correct answer: D

Hope this helps!

-Jay
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by swerve » Tue Jul 31, 2018 10:10 am
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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by [email protected] » Tue Jul 31, 2018 6:12 pm
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: D

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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by Scott@TargetTestPrep » Fri Aug 10, 2018 5:55 pm
BTGmoderatorDC 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 to do this.

Answer: D

Scott Woodbury-Stewart
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by Brent@GMATPrepNow » Sat Aug 11, 2018 5:14 am
BTGmoderatorDC 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
Once we recognize that we can achieve 5 points by drawing 3 Low cards and 1 High card, then it really comes down to determining the number of ways to rearrange 3 L's and 1 H.
One option is to simply list the arrangements.
Alternatively, we can use the MISSISSIPPI rule, which says:
If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
---------ONTO THE QUESTION---------------------------

Let's calculate the number of arrangements of the letters in LLLH:
There are 4 letters in total
There are 3 identical L's
So, the total number of possible arrangements = 4!/(3!)
= (4)(3)(2)(1)/(3)(2)(1)
= 4

Answer: D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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