## anonymous one year ago Probability: A balanced die is tossed six times, and the number on the uppermost face is recorded each time. What is the probability that the numbers recorded are 1, 2, 3, 4, 5, and 6 in any order?

1. anonymous

I just need help setting it up. I've got the computational power.

2. Vocaloid

intuitively I'd say 6!/(6^6)

3. Vocaloid

there are 6^6 total possibilities for one die being tossed 6 times since we want 1,2,3,4,5,6 in any order: for the first die, our possibilities are 1,2,3,4,5,6 for the second die, we've already used up one of the 6 options, so we're left with 5 for the third die, 4 choices, etc.

4. anonymous

Right, I do agree there's no replacement. Let me think about what you've written for a moment. Thank you

5. anonymous

In my textbook, the formula is n!/r!(n-r)!. It warns against making it n!/n^r because that doesn't work when there's no order.

6. anonymous

n being the sample and r being the selection from it

7. Vocaloid

hm..

8. anonymous

But if I do that, it's 6!/1!5! and that just turns out to be 6...which seems really small and an odd result

9. Vocaloid

that's what I was thinking, too. the formula should yield the same results

10. anonymous

What does 6 MEAN, then?

11. anonymous

What is that telling me?

12. Vocaloid

@satellite73 I think I'm missing something obvious here ;-;

13. dan815

hi okay so

14. anonymous

I really appreciate the effort here btw, this is the last question.

15. dan815

we have 6 numbers and 6 spots we want no repetition there are 6! of getting 1 to 6 in some order there are 6^6 possibilities for anything to happen therefore 6!/6^6 = chance of 1 to 6 in any order

16. anonymous

okay, let me look again at what my book says

17. anonymous

I guess my question is why doesn't the formula with no replacement and no order work here...

18. anonymous

Why do I have to make something up?

19. dan815

no replacements and no order eh

20. dan815

ya sure it works

21. anonymous

Okay, so how does n!/r!(n-r)! work here?

22. dan815

I am not fully sure what you mean by that formula though, i dont really know these formulas like that

23. dan815

okay so what u have there is (N choose R)

24. anonymous

Well I've got four of them for the variety of situations with order/without order and with/without replacement. I promise they're right.

25. anonymous

yep!

26. dan815

okay so for your formula, would you do 6 choose 6 then

27. anonymous

ohh maybe! Let me try

28. anonymous

No, then the formula gives me 6!/6!0! which is 1

29. anonymous

which kind of makes sense for the logic of "six choose six" I guess...I feel like I'm missing something big. I'm going to write down your improvised formula and then ask her before class tomorrow.

30. dan815

yep but wait there is nothing wrong with this

31. anonymous

32. dan815

you see how we didnt care about the order, now what are the total possibilitites where we dont care about the order again

33. anonymous

I don't know exactly what you're referring to, or saying. Please elaborate?

34. misty1212

HI!!

35. misty1212

there is replacement since you are tossing six different dice you do not use up a number once it is rolled

36. anonymous

oh right.

37. dan815

6^6 is how many total possibilties there are where arrangement matters so 6^6/6! so u see how if u do 1/6^6/6! = 6!/6^6 we are abck the first solution again

38. anonymous

The formula for no order and replacement is (n+r-1)!/r!(n-1)!

39. anonymous

I guess maybe I'm not clear on what n and r are. I think n is 6, but is r 1 or 6?

40. anonymous

ah misty left. nevermind

41. anonymous

I don't want to logic this out, dude. I want the forumlas to work. That's the point of having them. I won't get credit for my work if I pull something out of the blue. I will ask her tomorrow.

42. dan815

okay

43. anonymous

Ah it doesn't even matter if there's replacement, you get the same answer.

44. anonymous

im not im middle school......XD tf im in college

45. anonymous

bye now

46. anonymous

Kinda yeah XD

47. misty1212

my suggestion in doing probability is not to get married to any formula use whatever you need when you need it i wouldn't even use $\frac{n!}{k!(n-k)!}$ in that form ever

48. anonymous

Well you and my ph.d. professor would disagree, but I guess thanks.