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

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.

Answer: C.
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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

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

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

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!

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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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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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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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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Hi bessiebeardsley

Welcome to GMAT CLUB!!!

Please state your confusion clearly so that I can help you with them.

Can you please elaborate on this approach

Posted from my mobile device

Hi Niha01,

Five hands in general, could have any of the following outcomes:

Case 1) All 5 wins
Case 2) 4 wins and 1 loss
Case 3) 3 wins and 2 losses
Case 4) 2 wins and 3 losses
Case 5) 1 win and 4 losses
Case 6) All 5 losses

In the given question, we know there are 3 wins and 9 losses. Hence, the first five hands cannot have more than three wins. Hence cases 1) and 2) are not possible while any of the cases 3), 4), 5) or 6) are possible.

Now, there are two approaches to solve this question.

1st Approach:

Count the outcomes of the cases that are possible which is:
Case 3) 5C3 = 10
Case 4) 5C2 = 10
Case 5) 5C1 = 5
Case 6) 5C0 = 1
So, total possible outcomes = 10 + 10 + 5 + 1 = 26

2nd Approach: Subtract the outcomes of the cases that are not possible from the total no. of outcomes.

Case 1) 5C5 = 1
Case 2) 5C4 = 5

These two cases have to be subtracted from total no. of cases.

But how to find total no. of cases?

Each of the five hands has 2 possibilities - win or loss.

So five hands will have 2 x 2 x 2 x 2 x 2 = 32 possible outcomes.

So we subtract cases 1) and 2) from the total i.e. 32 - 1 - 5 = 26.

I hope this is understood. (Kudos if you have.)

Choice of which approach to use depends on which calculation will be lesser. If cases to be counted are less, use 1st approach. If cases to be excluded are less, use 2nd approach.

Hope this helped.

2 ways to finish every round.
So, there must be 2^5=32 ways to finish 5 rounds.

The possibility of
win each of the 5 rounds WWWWW= 1
win 4 rounds= 5C4= 5

He can win up to 3 rounds, gather $300+$110(the amount he started with)=$410 and lose the rest 12-3=9, thus 9*10=$90 still ends up with 410-90=$320

So, total ways= 2^5-(1+5)=26

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