Question

Three 10-sided dice, each numbered 1 through 10, are rolled. One die is green, one die is blue, and one die is red. Kareem rolls the three dice and writes down the numbers showing on the green die, the blue die, and the red die, in that order. There are 1000 different ordered triples that Kareem could have written down. For how many of those triples is the number on the blue die strictly between the numbers on the red and green dice?

Answer 1

Possible rolls with each die:
r1 r2 r3
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 8 8
9 9 9
10 10 10

Total number of different ordered triples: 10 x 10 x 10 = 1000

Answer 2

Looking at the response above, you see that the problem's statement that threre are 1000 different ordered triples is correct.

Now we need ordered triples in which the second number is between the first two.

Let's call the three rolls, r1, r2, and r3.

How many ordered triples are there where
r1 < r2 < r3?

If the first die roll is a 1, the second is 2, the third can be 3 through 10. That is 8 ordered triples.
If the first roll is a 1, the second is a 3, the third one can be 4, through 10. That is 7 more ordered triples.
This goes on until you have 1, 9, 10.
This gives you a total of 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 ordered triples.

Then you go on to 2, 3, 4, followed by 2, 3, 5, etc.
7 + 6 + 5 + 4 + 3 + 2 + 1 = 28

Then starting with 3, 4, n:
6 + 5 + 4 + 3 + 2 + 1 = 21

Then 4, 5, n:
5 + 4 + 3 + 2 + 1 = 15

Then 5, 6, n:
4 + 3 + 2 + 1 = 10

Then 6, 7, n:
3 + 2 + 1 = 6

Then 7, 8, n
2 + 1 = 3

Finally 8, 9, 10:
1

When you add all these nunmbers of ordered triples up you get
36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 = 120


Now if you add the ordered triples for which
r1 > r2 > r3,
you get another 120 orderred triples.

The answer is there are 240 ordered triples in which the middle number is strictly between the first and third numbers.