Behind The Times

@venda said
O.k Joe,I am really fascinated by how these things are done but (only)if you have the time and patience could you one day explain in simple terms where some of the numbers come from. A pm will do if you don't want to clutter up the thread.
I get the 11!
I get the 8!
I get the 5!
Where I get lost is the (9,2) (7,2) and the 8! 7! 6!

C(n,k) is the number of ways of choosing "k" objects out of "n" objects.

its equal to

C(n,k) = n!/( k! (n-k)! )

Is that the issue we should first go deeper with?

@bigdoggproblem said
But is it really an error, if I noticed it was happening and adjusted for it?

Your way is more elegant. I just didn't want to break that 11! apart, for some reason.

Error wasn't the best term. You get the correct answer.

Its a technicality on how probability is defined.

Probability = Favorable outcomes over All outcomes.

Your numerator and denominator don't hold true to the definition. Both of these things are overcounting the actual number of ways of each case ( by the same factor ).

AAHYSPPEERT is indistinguishable from AAHYSPPEERT, etc..., so we shouldn't be counting it all those extra times.

@joe-shmo said
Error wasn't the best term. You get the correct answer.

Its a technicality on how probability is defined.

Probability = Favorable outcomes over All outcomes.

Your numerator and denominator don't hold true to the definition. Both of these things are overcounting the actual number of ways of each case ( by the same factor ).

AAHYSPPEERT is indistinguishable from AAHYSPPEERT, etc..., so we shouldn't be counting it all those extra times.

I am unconvinced by this argument. I could just as easily argue that you arbitrarily chose to treat those cases as one single case, to simplify the math (not that there's anything wrong with that).



Are we justified in treating the two Rooks as if they were the same?

@bigdoggproblem said
I am unconvinced by this argument. I could just as easily argue that you arbitrarily chose to treat those cases as one single case, to simplify the math (not that there's anything wrong with that).

[fen]8/8/8/8/8/8/8/R3K2R[/fen]

Are we justified in treating the two Rooks as if they were the same?

I don't wan't to argue about it ( Its not like I'm your math teacher ). I said its a technicality. I am curious though,; could you write a computer program to distinguish between the two strings? What information between them computationally differentiable?

@joe-shmo
I'd have to add information to differentiate one P from another, e.g. label them p1 and p2.

@bigdoggproblem said
@joe-shmo
I'd have to add information to differentiate one P from another, e.g. label them p1 and p2.

Exactly. Without adding information, the strings are computationally indistinguishable. They add no new information to the set of all possible strings.

"p1p2" is not informationally identical to "pp"

You've added 349,272,200 "nothings to the set of all possible strings.

@joe-shmo
I thought you didn't want to argue. Hmm.

@bigdoggproblem said
@joe-shmo
I thought you didn't want to argue. Hmm.

I said I didn't want to argue about it, I didn't say I wouldn't!

@joe-shmo
😁

@bigdoggproblem said
@joe-shmo
😁

I'm sure you've heard that arguing with an Engineer is a lot like wrestling with a pig in the mud... After a couple of hours you realize the pig likes it!

😆

@joe-shmo
Speaking of engineers...I put this problem on the white board at work.

We'll see if any of ours can crack it.

(So far, the accountants have drawn a blank.)

@bigdoggproblem said
@joe-shmo
Speaking of engineers...I put this problem on the white board at work.

We'll see if any of ours can crack it.

(So far, the accountants have drawn a blank.)

Accountants...pfff. "Count" is literally in their job title...what do they do for 4 years?

😉

@joe-shmo said
C(n,k) is the number of ways of choosing "k" objects out of "n" objects.

its equal to

C(n,k) = n!/( k! (n-k)! )

Is that the issue we should first go deeper with?

I have that somwhere thanks.
It's the standard combinations formula, usually quoted as n!/(n-r!)*r!)
I'll have another think while you spar with big dogg!!

@bigdoggproblem said
Yes, I wrote a program to solve this problem.

I am not sure how I want to share it. It's written in Python. I would post the code here, but RHP removes the spacing.

Doesn't matter mate.
I know about the python programming tool.
I plan to look at it more closely sometime.

@joe-shmo said
C(n,k) is the number of ways of choosing "k" objects out of "n" objects.

its equal to

C(n,k) = n!/( k! (n-k)! )

Is that the issue we should first go deeper with?

O.k Joe ,I've got a bit further
Your formula C(11,2)*C(9,2)*C(7,2)*5! = 55*36*21*5! Is actually the same calculation as 11!/2!2!2!.
I just need to look at the probability part now!
As an aside,I looked in an book I once compiled when I encountered this sort of thing years ago and I had written::-
The way to find any number of objects "p" of one kind and "q" of another with the rest all different = (p+1)(q+1)2^n-p-q-1
I tried that in a spreadsheet but couldn't get the same answer.
I suppose I may have written it down wrong.