Problem Solving

The question can be found on www.manhattangmat.com

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

x won and y lost
x+y = 12
100x-10y= 210
=> x= 3, y =9
So, in first five hands there can be following combinations
w 3, L 2
W 2, L 3
W 1, L 4
W 0, L 5

In first case, 3 w can be arranged in 5 hands in 5C3 i.e. 10 ways
Similarly 2nd case 5C2= 10
3rd case 5C1= 5
4th case 5C0= 1
So, the number of possibilities is 10+10+5+1 = 26

Answer should be C (I think so)

I agree with nkumar it should be 26

32 - 1 ( all 5 wins) - 5 ( 4 wins & 1 loss) =26

12 attemps = 210$ so he has to win 3 times atleast to take the amt above 200 & loose 9 times to bring it back to 210

26 for me too ..