Author 
Message 
TAGS:

Hide Tags

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

A gambler began playing blackjack with $110 in chips. After [#permalink]
Show Tags
22 Mar 2005, 04:01
7
This post received KUDOS
21
This post was BOOKMARKED
Question Stats:
61% (03:38) correct
39% (02:45) wrong based on 512 sessions
HideShow timer Statistics
A gambler began playing blackjack with $110 in chips. After exactly 12 hands, he left the table with $320 in chips, having won some hands and lost others. Each win earned $100 and each loss cost $10. How many possible outcomes were there for the first 5 hands he played? (For example, won the first hand, lost the second, etc.) (A) 10 (B) 18 (C) 26 (D) 32 (E) 64
Official Answer and Stats are available only to registered users. Register/ Login.



Senior Manager
Joined: 19 Feb 2005
Posts: 486
Location: Milan Italy

1
This post received KUDOS
I guess! (C)
well the algebra first > 110 (starting cash)+100x (n of wins)10*(12x)=320 > solve and get x=3
if wins=3/12
if losses= 9/12
there can be max 3 wins out of 5 hands
alas, I'm not sure about how to turn out the outcomes
I think
5c0+5c1+5c2+5c3=26
that is 0 wins/1 win/2 wins/3 wins



Senior Manager
Joined: 19 Nov 2004
Posts: 281
Location: Germany

I got the algebra bit correct too: 3 Wins and 9 losses.
I'm unsure on how to go about the arranging part. thearch, I guess your approach looks fine and C should be the answer.
Good question, ywilfred!



Manager
Joined: 15 Feb 2005
Posts: 246
Location: Rockville

i would have to go with C also because he can only have 3 wins and 9losses in 12 rounds
so in 5 turns the possibilities are limited to getting no wins, 1 win, 2 wins and all 3 wins.
26



VP
Joined: 18 Nov 2004
Posts: 1433

3
This post received KUDOS
4
This post was BOOKMARKED
"C"
If X is number of wins and Y num of losses then
100X  10Y = 210
10XY = 21......only when Y = 9 and X = 3 it satisfies....so we have 3 wins and 9 losses.
for first 5, we can have the following:
0 wins 5 losses = 1 way
1 win and 4 losses = 5C1 = 5
2 wins and 3 losses = 5C2 = 10
3 wins and 2 losses = 5C3 = 10
Add total ways = 26 ways



SVP
Joined: 03 Jan 2005
Posts: 2233

Good job thearch. You can also do it this way: 2^55(four wins)1(five wins)=26.



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15964

Re: A gambler began playing blackjack with $110 in chips. After [#permalink]
Show Tags
24 Feb 2014, 03:18
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Intern
Joined: 18 Sep 2013
Posts: 7

What about arrangement of these outcomes, don't we have to arrange their sequence? I applied permutation instead of combination, got the wrong answer. Kindly help Thanks!



Math Expert
Joined: 02 Sep 2009
Posts: 39666

9
This post received KUDOS
Expert's post
3
This post was BOOKMARKED
ShantnuMathuria wrote: What about arrangement of these outcomes, don't we have to arrange their sequence? I applied permutation instead of combination, got the wrong answer. Kindly help Thanks! A gambler began playing blackjack with $110 in chips. After exactly 12 hands, he left the table with $320 in chips, having won some hands and lost others. Each win earned $100 and each loss cost $10. How many possible outcomes were there for the first 5 hands he played? (For example, won the first hand, lost the second, etc.) (A) 10 (B) 18 (C) 26 (D) 32 (E) 64 The gambler started with $110 and left with $320, thus he/she in 12 hands won $320  $110 = $210: 100W  10L = 210; 100W  10(12W) = 210 (since Wins + Loss = 12) > W = 3. So, we have that out of 12 hands the gambler won 3 hands and lost 9. For the first 5 hands played there could be the following outcomes: WWWLL > 5!/(3!2!) = 10 ways this to occur (for example, WWWLL, WWLWL, WLWWL, ...); WWLLL > 5!/(3!2!) = 10 ways this to occur; WLLLL > 5!/(4!1!) = 5 ways this to occur; LLLLL > only 1 way this to occur. Total = 10 + 10 + 5 + 1 = 26. Answer: C.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 22 Feb 2014
Posts: 2

Re: A gambler began playing blackjack with $110 in chips. After [#permalink]
Show Tags
01 Apr 2014, 20:44
Can someone explain where I can learn more about this: WWWLL > 5!/(3!2!) = 10 Why do you divide here. I think i get the logic, but how do you know to choose 3! and 2!, and how can I know when to do this.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7440
Location: Pune, India

Re: A gambler began playing blackjack with $110 in chips. After [#permalink]
Show Tags
01 Apr 2014, 21:11
frenchwr wrote: Can someone explain where I can learn more about this: WWWLL > 5!/(3!2!) = 10 Why do you divide here. I think i get the logic, but how do you know to choose 3! and 2!, and how can I know when to do this. You arrange 5 distinct objects in 5! ways. But if some of them are identical, you need to divide the total arrangements by the factorial of that number: Say you have total n objects out of which m are identical. Total number of arrangements = n!/m! e.g. Out of 5 objects, if 2 are identical, number of arrangements = 5!/2! (because we don't have as many arrangements as before now.) Say 5 objects are A, B, C, D and D. There are 2 identical Ds. 5! gives the arrangements of 5 distinct objects(e.g. ABCDE, ABCED are two diff arrangements) but if two letters are same, ABCDD is same as ABCDD (we flipped the D with the other D). Hence the number of arrangements are half in this case: 5!/2! Similarly, if you have 5 letters such that three of them are same and another 2 are same, the number of arrangements is given by 5!/(3!*2!) as is the case with WWWLL. Our Combinatorics book discusses this concept as well as other GMAT relevant concepts in detail. You can take a look at it here: http://www.amazon.com/VeritasPrepStat ... ds=veritas
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Intern
Joined: 10 May 2015
Posts: 30

Re: A gambler began playing blackjack with $110 in chips. After [#permalink]
Show Tags
24 May 2015, 10:13
Guys...............I have a strange doubt about this..okay we so know there are 3 wins and 9 losses..and to have 5 hands..you could select 3 wins and 2 losses say..now why can't we just do 9c2(2losses out of 9)* 3c3(3wins out of 3).If we say it's identical then you can divide it by 3 ! and 2! respectively.However,that's still miserably wrong..I know it's wrong but can't figure out why? Please help!!!



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7440
Location: Pune, India

Re: A gambler began playing blackjack with $110 in chips. After [#permalink]
Show Tags
24 May 2015, 19:55
davesinger786 wrote: Guys...............I have a strange doubt about this..okay we so know there are 3 wins and 9 losses..and to have 5 hands..you could select 3 wins and 2 losses say..now why can't we just do 9c2(2losses out of 9)* 3c3(3wins out of 3).If we say it's identical then you can divide it by 3 ! and 2! respectively.However,that's still miserably wrong..I know it's wrong but can't figure out why? Please help!!! You cannot select losses out of losses  they are all just losses. You can select hands to which you will allot losses since the hands are distinct  first hand, second hand .. till 12th hand. For example, if we have to give 3 wins and 2 losses to 5 hands, we can select the 2 hands to which we will give losses. We can do this in 5C2 ways = 10 ways. The other 3 hands will automatically be left with wins. This is another way of doing what Bunuel did above. Similarly, to give 3 losses we select 3 hands out of 5 in 5C3 ways = 10 ways To give 4 losses, we select 4 hands out of 5 in 5C4 = 5 ways To give 5 losses, we select 5 hands out of 5 in 5C5 = 1 way Total = 10+10+5+1 = 26
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Intern
Joined: 10 May 2015
Posts: 30

Re: A gambler began playing blackjack with $110 in chips. After [#permalink]
Show Tags
24 May 2015, 20:14
Thank you Karishma for your reply. So ,in this case do you mean to say that since they're all identical,we can't select losses out of losses?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7440
Location: Pune, India

Re: A gambler began playing blackjack with $110 in chips. After [#permalink]
Show Tags
24 May 2015, 20:43
davesinger786 wrote: Thank you Karishma for your reply. So ,in this case do you mean to say that since they're all identical,we can't select losses out of losses? Yes, think of it this way: If you have 12 different houses and you have to paint 5 of them  3 red and 2 yellow, can you select red out of red? You must select the 3 houses out of 5 which you will paint red.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Intern
Joined: 10 May 2015
Posts: 30

Re: A gambler began playing blackjack with $110 in chips. After [#permalink]
Show Tags
25 May 2015, 20:13
ah I see your point.Thank you for that,Karishma!



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15964

Re: A gambler began playing blackjack with $110 in chips. After [#permalink]
Show Tags
13 Jun 2016, 08:20
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: A gambler began playing blackjack with $110 in chips. After
[#permalink]
13 Jun 2016, 08:20







