A standard six-sided die is rolled 7 times. You are told that among the rolls, there was one 1, two 2's, and three 4's. How many possible sequences of rolls could there have been? (For example, 5, 1, 2, 2, 4, 4, 4 is one possible sequence. 4, 1, 4, 4, 2, 1, 1 is one
+2668 Let's divide this problem up into cases:
Case 1 (2 1's, 2 2's, 3 4's) - There are \({7! \over 2! 2! 3!} = 210\) cases.
Case 2 (1 1's, 3 2's, 3 4's) - There are \({7! \over 3! 3!} = 140\) cases.
Case 3 (1 1's, 2 2's, 4 4's) - There are \({7! \over 2! 4!} = 105\) cases.
Case 3 (1 1's, 2 2's, 3 4's, 1 3, 5, or 6) - There are \({7! \over 2! 3!} \times 3 = 1260\) cases.
So, there are \(210 + 140 + 105 + 1260 = \color{brown}\boxed{1,715}\) sequences.
