q35

[Saumyojit]

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?

 

[Dr.Peterson]

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?

How do you know you're wrong? Once again, if there are choices given, they are part of the problem; and if you saw the "correct" answer, it might be wrong, so we need to see that, too.

But you're wrong at least because you didn't finish; and also because that approach will take a long time; and also because the calculations you did are wrong (to the extent that I can tell what you mean). Once you place the 1 and the 6, there are always 4! ways to place the rest.

On this other hand, have you considered that 1 will always be either before or after 6? How many ways are there to have 1 after 6?

 

[sky2rain]

How do you know you're wrong? Once again, if there are choices given, they are part of the problem; and if you saw the "correct" answer, it might be wrong, so we need to see that, too.

But you're wrong at least because you didn't finish; and also because that approach will take a long time; and also because the calculations you did are wrong (to the extent that I can tell what you mean). Once you place the 1 and the 6, there are always 4! ways to place the rest.

On this other hand, have you considered that 1 will always be either before or after 6? How many ways are there to have 1 after 6?

Equal events for 1 before 6 and after 6, so p6,6 / 2. If using the direct approach, these are the following formats for 1 before 6: 16||||, |16|||… 1|6|||,|1|6||…1||6||,|1||6|…1|||6|,|1|||6 and 1||||6, (5+4+3+2+1)*P4,4 ways. Not sure if this is right.

 

[Dr.Peterson]

Equal events for 1 before 6 and after 6, so p6,6 / 2. If using the direct approach, these are the following formats for 1 before 6: 16||||, |16|||… 1|6|||,|1|6||…1||6||,|1||6|…1|||6|,|1|||6 and 1||||6, (5+4+3+2+1)*P4,4 ways. Not sure if this is right.

Both answers are right (and, in fact, equal), though it's hard to follow your thinking on the second method.

I would say this: Considering only the placement of the 1 and the 6, if 1 is in the first position, there are 5 places after it to put the 6 (1 _ _ _ _ _); if 1 is in the second position, there are 4 places to put the 6 (x 1 _ _ _ _), and so on down to 1 being in the last place, with no places to put the 6. Once those are placed (e.g. _1 _ _ 6 _), the other digits can be put in the remaining 4 places in 4! ways. So the total is, as you say, ((5+4+3+2+1+0)*4!.