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A gambler began playing blackjack with $110 in chips. After [#permalink]
22 Mar 2005, 03:01

3

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A

B

C

D

E

Difficulty:

85% (hard)

Question Stats:

59% (03:40) correct
41% (03:07) wrong based on 159 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.)

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

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

Re: A gambler began playing blackjack with $110 in chips. After [#permalink]
24 Feb 2014, 02:18

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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

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.

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.

Re: A gambler began playing blackjack with $110 in chips. After [#permalink]
01 Apr 2014, 20:11

1

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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.

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