Three numbers are randomly selected, with replacement, from the set of integers {0, 1, 2, 3, 4, 5, 6, 7}. If the first number selected is w, the second number selected is x, and the third number is y, what is the probability that w < x < y ? Choose the correct option:  A) 5/32  B) 7/64  C) 9/64  D) 53/512  E) 81/512
The number of ways to select w, x, and y is:   8*8*8= 
<gadget id="calculator">  8*8*8</gadget>
<output>512</output>
512   The number of ways to choose three different numbers from the set is 8C3=56.   There are six permutations for each group of three different numbers, but only one permutation satisfies w &lt; x &lt; y.   P(w &lt; x &lt; y) = 56/512 = 7/64   The answer is 7/64.
<result>B</result>
Three numbers are randomly selected, with replacement, from the set of integers {0, 1, 2, 3, 4, 5, 6, 7}. If the first number selected is w, the second number selected is x, and the third number is y, what is the probability that w < x < y ?