3 numbers are randomly selected, with replacement, from the set of integers {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. 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 answer:
A) 3/40
B) 28/243
C) 3/25
D) 33/100
E) 64/125
Solution:   Total possible ways = 10*10*10= 
<gadget id="calculator">10*10*10</gadget>
<output>1_000</output>
1000   Case1: w=0.   if x = 1, then y can have 8 ways.   if x =2 , then y can have 7 ways and so on.   So, no. of ways = 8+7+..+1   Case2 : w=1.   if x = 2, then y can have 7 ways.   if x =3 , then y can have 6 ways and so on.   So, no. of ways = 7+6 +..+1   There will be 8 cases in this way and they will follow a pattern which looks like this:   No. of ways for Case1 : 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8   No. of ways for Case2 : 1 + 2 + 3 + 4 + 5 + 6 + 7   No. of ways for Case3 : 1 + 2 + 3 + 4 + 5 + 6   No. of ways for Case4 : 1 + 2 + 3 + 4 + 5   No. of ways for Case5 : 1 + 2 + 3 + 4   No. of ways for Case6 : 1 + 2 + 3   No. of ways for Case7 : 1 + 2   No. of ways for Case8 : 1   ____________________________________________________________   Total number of ways : 8(1) + 7(2) + 6(3) + 5(4) + 4(5) + 3(6) + 2(7) + 1(8) = 2(4(5) + 3(6) + 2(7) + 1(8)) = 2(60) = 120   Therefore, ans is 120/1000 = 3/25.   Option c.
<result>C</result>
3 numbers are randomly selected, with replacement, from the set of integers {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. 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 ?