statistics-need help, my brain has crashed!

[lemonjelly 0]

Being a left hander, The creative side of my brain is prominent..
This is doing my nut in!

Problem:
I have 14 selections to make, i can enter 1, 2 or 3
I know for a fact that 6 of them will be 1

how many different permutations are there to cover all eventualities??

Oh, And I've just found out that I'm clean out of potatoes

*************************************************
RED LIGHTS & OFF LANDINGS ARE JUST MY THANG
http://www.redlightrob.co.uk

[CrazyIvan 0]

24

__________________________________________
Blue Skies and May the Force be with you.

[lemonjelly 0]

are you sure?

*************************************************
RED LIGHTS & OFF LANDINGS ARE JUST MY THANG
http://www.redlightrob.co.uk

[lemonjelly 0]

That answer was as good as a repost

*************************************************
RED LIGHTS & OFF LANDINGS ARE JUST MY THANG
http://www.redlightrob.co.uk

[lemonjelly 0]

how many statisticians does it take to change a lightbulb?

*************************************************
RED LIGHTS & OFF LANDINGS ARE JUST MY THANG
http://www.redlightrob.co.uk

[katzurki 0]

14 selections, 6 of which are fixed. That leaves 8 variable. Each can be 1, or 2, or 3 -- three total. Then it's 3^8 = 6561

That is, unless you want to count the order in which you make these selections, in which case the number of the permutations increases significantly.

[Lostinspace 0]

Permutations of where the 1 can sit:

n_t=total slots=14
n_1=number of 1’s=6

# of permutaions= (n_t)!/[(n_1)!*(n_t-n_1)!]

Number of why the the 2’s and 3;s can sit in their slots: 2^8

So you number of permutations is:
(2^8)* (n_t)!/[(n_1)!*(n_t-n_1)!]

(2^8)* (14)!/[(6)!*(8)!]=768,768

768,768

I may not be right. That would be my final answer.

That will be a dollar please.

Now if this were a game, and I had a 50 percent chance of being correct; but I had to pay you every time I tried to school you, but when I finally did school you, you would pay me 2^n, where n is the number of tries I have had and I quit the game when I succeed, how much should you charge me for this game to be fair?