On a game show, a contestant is given three keys, each of which opens exactly one of three identical boxes. The first box contains $2, the second $200, and the third $2000. The contestant assigns each key to one of the boxes and wins the amount of money contained in any box that is opened by the key assigned to it. What is the probability that a contestant will win more than $2000?	Choose the correct choice from the following choices:
A) 1/9
B) 1/8
C) 1/3
D) 1/6
E) 1/2
Let's call the boxes that contain $2, $200, and $2000, respectively, Box A, Box B, Box C. These are opened, respectively, by Key A, Key B, and Key C.   We want to know the probability of winning more than $2000. Notice that if the distribution of keys is:   Box A = Key B   Box B = Key A   Box C = Key C   then the contestant wins exactly $2000, not more than $2000. The only configuration that leads to winning more than $2000 is:   Box A = Key A   Box B = Key B   Box C = Key C   i.e., getting all three keys correct. That's the only way to be more than $2000. So, really, the question can be rephrased: what is the probability of guessing the order of keys so that each key matches the correct box?   Well, for a set of three items, the number of possible permutations is 3! = 3*2*1= 
<gadget id="calculator">3*2*1</gadget>
<output>6</output>
6.   Of those 6 possible permutations, only one of them leads to all three keys being paired with the right box. So, the answer is   Probability= 1/6,   Answer = 1/6
<result>D</result>
On a game show, a contestant is given three keys, each of which opens exactly one of three identical boxes. The first box contains $2, the second $200, and the third $2000. The contestant assigns each key to one of the boxes and wins the amount of money contained in any box that is opened by the key assigned to it. What is the probability that a contestant will win more than $2000?