How many nonrepeating linear number sequences can you get from the throw of six dice?

Question

Each die is numbered 1 to 6. Each throw of six dice will reveal a potentially linear (imagine they fall in a straight line) sequence of six numbers, often, but not always, repeated numbers, ie, 1, 1, 1, 3, 4, 5, or 2, 2, 3, 3, 1, 5. Equally you might throw a nonrepeating linear number sequence, such as,1, 2, 3, 4, 5, 6, or 6, 1, 4, 5, 2, 3. My question is simply this: how many nonrepeating linear number sequences is it possible to throw with six dice. This is not a chance question, ie what is the chance of throwing a nonrepeating linear number sequence, but simply how many nonrepeating linear number sequences can you throw with six dice? How would you calculate this?

Answer 1

This sounds like a combinatorics question which asks you to calculate the number of possible permutations for a given set of something. You should be able to figure out what the answer is by checking out the example provided on the link below.

Answer 2

Factorial 6, or 6! = 6*5*4*3*2*1 = 720

Answer 3

i'm assuming you propose getting a unusual selection the two time. The threat is 3/4 or seventy 5%. you have a 50% risk the two rolls -- combining the possibilities supplies seventy 5% risk for hitting one extraordinary selection in 2 rolls.

Answer 4

6x5x4x3x2x1=720
The first die has all 6 numbers available. The second die has all 6 minus the number from the first die. etc etc

Answer 5

More homework, or is it an exam?

Answer 6

Thanks to the responses above I know the answer and how to work it out......

Answer 7

UUHhhhhhh...............