sukhman 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
Let W = the number of wins and L = the number of losses.
Since there are a total of 12 hands, we get:
W+L = 12.
Since he starts with $110 and leaves with $320, his total earnings = 320-110 = 210.
Since each win adds to his earnings $100, and each loss decreases his earnings by $10, we get:
100W - 10L = 210.
10W - L = 21.
Adding together the two equations, we get:
(W+L) + (10W-L) = 12+21
11W= 33.
W=3
Since W+L=12, L=9.
Thus, the 12 hands are composed of 3 wins and 9 losses:
WWWLLLLLLLLL.
The first 5 hands may be composed of up to 3 wins:
WWWLL
WWLLL
WLLLL
LLLLL.
To determine the number of options for the first 5 hands, we must count the number of ways to arrange the 5 letters in each of the four cases above.
For example, case 3 yields the following options for the first 5 hands:
WLLLL, LWLLL, LLWLL, LLLWL, LLLLW.
The number of ways to arrange 5 DISTINCT elements = 5!.
But the elements in the 4 cases above are not all distinct.
When an arrangement includes IDENTICAL elements, we must divide by the number of ways the identical elements can be arranged.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Case 1: WWWLL
Here, we must divide by 3! to account for the 3 identical W's and by 2! to account for the 2 identical L's:
5!/(3!2!) = 10.
Case 2: WWLLL
Here, we must divide by 2! to account for the 2 identical W's and by 3! to account for the 3 identical L's:
5!/(2!3!) = 10.
Case 3: WLLLL
Here, we must divide by 4! to account for the 4 identical L's:
5!/4! = 5.
Case 4: LLLLL
Here, we must divide by 5! to account for the 5 identical L's:
5!/5! = 1.
Total ways = 10+10+5+1 = 26.
The correct answer is
C.
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