Plz help
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 possible sequence.)
(1, 1, 2, 2, 4, 4, 4)=7!/3!2!2! =210 permutations
(1, 2, 2, 2, 4, 4, 4)=7!/3!3! =140 p
(1, 2, 2, 3, 4, 4, 4)=7!/3!2! =420 p
(1, 2, 2, 4, 4, 4, 4)=7!/4!2! =105 p
(1, 2, 2, 4, 4, 4, 5)=7!/3!2! =420 p
(1, 2, 2, 4, 4, 4, 6)=7!/3!2! =420 p
Total =210 + 140 + 420 + 105 + 420 + 420 =1,715 possible sequences of rolls
