+8 
There are 3 combinations as follows:
(1, 5, 5, 6, 6) ==5!/2!2! ==30 permutations
(2, 3, 5, 5, 6) ==5!/2! == 60 permutations
(3, 3, 4, 5, 5) ==5! /2!2! ==30 permutations
Total ==30 + 60 + 30 == 120 different sequences of rolls.
+8 Ahhh yes, I failed to consider the combinations with 1s in them. Thank you so much, and I now understand.
To find the number of different sequences of rolls that could have resulted in a product of 900, we need to consider the prime factorization of 900.
The prime factorization of 900 is:
900 = 2² * 3² * 5²
We can distribute these prime factors among the five rolls of the die. Since the order of the rolls matters, we can think of each roll as a "slot" where we can place one of the prime factors.
Let's consider the number of ways to distribute the prime factor 2² = 4 among the five rolls. We can have:
4 in the first roll, 0 in the second, third, fourth, and fifth rolls.
3 in the first roll, 1 in the second, 0 in the third, fourth, and fifth rolls.
2 in the first roll, 2 in the second and 0 in the third, fourth, and fifth rolls.
2 in the first roll, 1 in the second, 1 in the third, and 0 in the fourth and fifth rolls.
1 in the first roll, 1 in the second, 1 in the third, and 2 in the fourth and fifth rolls.
Now, let's consider the number of ways to distribute the prime factor 3² = 9 among the remaining slots. We can have:
9 in the first roll, 0 in the second, third, fourth, and fifth rolls.
8 in the first roll, 1 in the second, 0 in the third, fourth, and fifth rolls.
7 in the first roll, 2 in the second and 0 in the third, fourth, and fifth rolls.
6 in the first roll, 3 in the second, 0 in the third, fourth, and fifth rolls.
6 in the first roll, 2 in the second, 1 in the third, and 0 in the fourth and fifth rolls.
5 in the first roll, 4 in the second, 0 in the third, fourth, and fifth rolls.
5 in the first roll, 3 in the second, 1 in the third, and 0 in the fourth and fifth rolls.
4 in the first roll, 4 in the second, 1 in the third, and 0 in the fourth and fifth rolls.
4 in the first roll, 3 in the second, 2 in the third, and 0 in the fourth and fifth rolls.
3 in the first roll, 3 in the second, 2 in the third, and 1 in the fourth and fifth rolls.
Finally, for the remaining prime factor 5² = 25, we have only one possibility:
25 in the first roll, 0 in the second, third, fourth, and fifth rolls.
By multiplying the number of possibilities for each prime factor, we find the total number of different sequences of rolls:
5 * 10 * 1 = 50
Therefore, there are 50 different sequences of rolls that could have resulted in a product of 900.
