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# A gambler began playing blackjack with \$110 in chips. After

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A gambler began playing blackjack with \$110 in chips. After [#permalink]

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22 Mar 2005, 04:01
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Question Stats:

61% (02:30) correct 39% (02:29) wrong based on 484 sessions

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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

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22 Mar 2005, 05:25
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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

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22 Mar 2005, 06:08
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!

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22 Mar 2005, 09:49
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

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22 Mar 2005, 10:47
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"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

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25 Mar 2005, 11:04
Good job thearch. You can also do it this way: 2^5-5(four wins)-1(five wins)=26.

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11 Mar 2014, 10:28
What about arrangement of these outcomes, don't we have to arrange their sequence?
Kindly help

Thanks!

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11 Mar 2014, 10:59
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ShantnuMathuria wrote:
What about arrangement of these outcomes, don't we have to arrange their sequence?
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.

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Re: A gambler began playing blackjack with \$110 in chips. After [#permalink]

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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.

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Re: A gambler began playing blackjack with \$110 in chips. After [#permalink]

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01 Apr 2014, 21:11
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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
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Re: A gambler began playing blackjack with \$110 in chips. After [#permalink]

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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!!!

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Re: A gambler began playing blackjack with \$110 in chips. After [#permalink]

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24 May 2015, 19:55
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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
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Re: A gambler began playing blackjack with \$110 in chips. After [#permalink]

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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?

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Re: A gambler began playing blackjack with \$110 in chips. After [#permalink]

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24 May 2015, 20:43
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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.
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Re: A gambler began playing blackjack with \$110 in chips. After [#permalink]

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25 May 2015, 20:13
ah I see your point.Thank you for that,Karishma!

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A gambler began playing blackjack with \$110 in chips. After [#permalink]

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29 Oct 2017, 17:45
ywilfred wrote:
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

We can see there have to be 3 wins and 9 losses for the profit to be \$210 after 12 hands

He could have won all the three in the first 5 games or won 2 or won 1 or won none in the first 5 games.

So the number of possibilities are 5C3 + 5C2 + 5C1 +5C0=26
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A gambler began playing blackjack with \$110 in chips. After   [#permalink] 29 Oct 2017, 17:45
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