Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 04 May 2015, 16:18

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# A gambler began playing blackjack with $110 in chips. After  Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: GMAT Club Legend Joined: 07 Jul 2004 Posts: 5078 Location: Singapore Followers: 22 Kudos [?]: 184 [3] , given: 0 A gambler began playing blackjack with$110 in chips. After [#permalink]  22 Mar 2005, 03:01
3
KUDOS
2
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

59% (03:36) correct 41% (03:00) wrong based on 174 sessions
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
[Reveal] Spoiler: OA
Senior Manager
Joined: 19 Feb 2005
Posts: 487
Location: Milan Italy
Followers: 1

Kudos [?]: 11 [1] , given: 0

1
KUDOS
I guess! (C)
well the algebra first -> 110 (starting cash)+100x (n of wins)-10*(12-x)=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: 284
Location: Germany
Followers: 1

Kudos [?]: 13 [0], given: 0

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: 247
Location: Rockville
Followers: 1

Kudos [?]: 6 [0], given: 0

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: 1442
Followers: 2

Kudos [?]: 20 [2] , given: 0

2
KUDOS
2
This post was
BOOKMARKED
"C"

If X is number of wins and Y num of losses then

100X - 10Y = 210

10X-Y = 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: 2250
Followers: 13

Kudos [?]: 216 [0], given: 0

Good job thearch. You can also do it this way: 2^5-5(four wins)-1(five wins)=26.
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 4767
Followers: 296

Kudos [?]: 52 [0], given: 0

Re: A gambler began playing blackjack with $110 in chips. After [#permalink] 24 Feb 2014, 02:18 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. _________________ Intern Joined: 17 Sep 2013 Posts: 7 Followers: 0 Kudos [?]: 0 [0], given: 18 Re: [#permalink] 11 Mar 2014, 09:28 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: 27215 Followers: 4228 Kudos [?]: 41011 [4] , given: 5654 Re: Re: [#permalink] 11 Mar 2014, 09:59 4 This post received KUDOS Expert's post 1 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(12-W) = 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.

_________________
Intern
Joined: 22 Feb 2014
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: A gambler began playing blackjack with $110 in chips. After [#permalink] 01 Apr 2014, 19:44 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: 5453 Location: Pune, India Followers: 1335 Kudos [?]: 6783 [1] , given: 177 Re: A gambler began playing blackjack with$110 in chips. After [#permalink]  01 Apr 2014, 20:11
1
KUDOS
Expert's post
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/Veritas-Prep-Stat ... ds=veritas
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Veritas Prep GMAT course is coming to India. Enroll in our weeklong Immersion Course that starts March 29!

Veritas Prep Reviews

Re: A gambler began playing blackjack with $110 in chips. After [#permalink] 01 Apr 2014, 20:11 Similar topics Replies Last post Similar Topics: 2 The game of blackjack is played with a deck consisting of 6 27 Feb 2013, 03:54 6 A gambler bought$3,000 worth of chips at a casino in 7 25 Feb 2012, 21:23
1 In a casino, a gambler stacks a certain number of chips in 12 18 Jun 2010, 03:13
2 In a certain game played with red chips and blue chips, each 2 28 Jan 2007, 07:44
in a certain game played with red chips and blue chips, each 1 08 Oct 2006, 09:22
Display posts from previous: Sort by