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LOL jayzdd gets the medal for an INCORRECT answer ??? !!!
The answer is NINE!!
LOL

Heck, I'LL send a medal to jayzdd too!!!

I did not finish

Nice derangement formula. I did this by making a tree which is tedious

Hello to ganeshie8 :-)

Or you could have just copied my numbers :-)

|dw:1434277905759:dw|

some of these branches won't work, because we are left with a 4 in the last branch

and anyway, to continue with your ABCD method the ones remaining are:
BDAC
CADB
CDBA
DCAB

|dw:1434278137565:dw|

Thankyou . we came out with the same exact solution. Maybe your trial and error was similar?

And scrolling down, the scroll bar stops about half way down.

do*

Do you know how they derived that derangements formula?

Thats your website . Nice

oh :)

Heck I bet you never knew about "derangements" before tonight right?

I'll research this.

its nine I get it.....Thanks for the answer

If i follow that , that will explain the process?

Thats cool.
The probability of selecting a derangement of n items is 1/e as n->oo

Gonna get going jayzdd
Nice meeting you here.
c ya :-)

Thanks

Interesting, you found a recursive formula to produce derangements.

With some consideration whether n is even or odd

There might be a way to prove using advanced math techniques.