A die is rolled 6 times. 1, 2, 3, 4, 5 and 6 appear on consecutive throws of the die. How many ways are possible of having 1 before 6?
(1,2,3,4,5,6) Suppose six always come on the last turn then 5! = 120
(1,2,3,4,6,5) Suppose six always come on 2nd last turn, 4!=24 ways
(1,2,3,6,5,4) 3!= 6 ways....so on
Where am i wrong?
(1,2,3,4,5,6) Suppose six always come on the last turn then 5! = 120
(1,2,3,4,6,5) Suppose six always come on 2nd last turn, 4!=24 ways
(1,2,3,6,5,4) 3!= 6 ways....so on
Where am i wrong?