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GMAT Club Legend
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A gambler began playing blackjack with $110 in chips. After [#permalink]
 22 Mar 2005, 04:01
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 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?
(A) 10
(B) 18
(C) 26
(D) 32
(E) 64
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Director
Joined: 19 Feb 2005
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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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Senior Manager
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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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GMAT Club Legend
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I'll let on the OA in a couple more days.. see if any more members like to give a shot 
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Senior Manager
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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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SVP
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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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GMAT Club Legend
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You're all right. 26 (Choice C) is indeed the answer.
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Director
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sorry but I never studied combinations and perm at high school so it's a great result for me
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SVP
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Good job thearch. You can also do it this way: 2^5-5(four wins)-1(five wins)=26.
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close
