here is a brain buster..

[Semidevil]

ok, it's actually a homework questin....


so, how many ways can the letters in CINCINNATI be arranged so that no 2 I's are adjecent.

All I have is 10!/(2! * 3! * 3!) = 50400 total ways to be arranged....

so how do you add the limiting "no 2 I's be adjacent?

 

[nakedfrog]

I have a better question. Who gives a rat's arse?

 

[Shooters]

Isn't this just a counting problem?

Find the number of ways that 2 I's can be adjacent and then subtract it from the total number of ways you've already come up with.

 

[vshah]

treat the 2 Ls as an arbitrary symbol * and see how many times that is possible, then subtract




EDIT: this was thoguht up in 2 milliseconds and is probably incorrect.

-Vivan

 

[chuckywang]

CINCINNATI has

2 C's
3 I's
3 N's
1 A
1 T

for a total of A=10!/(2!*3!*3!*1!*1!) ways to arrange the letters.

Out of that many ways to arrange the letters, you have to figure out how many ways have 2 I's together.
The number of ways that can happen is:

B=(8!/(2!*1!*3!*1!*1!))*9 - (7!/(2!*3!*1!*1!))*8


Therefore, the total number of ways to arrange the letters so that no 2 I's are adjacent is

A-B.

 

[Evadman]

Depends on how you determine what is being arranged. There are 2 "n"'s next to each other. if they swap places, does that count as a new instance? It is still the same thing really. That would involve another step. Whatta say?

 

[Shooters]

Originally posted by: Evadman
Depends on how you determine what is being arranged. There are 2 "n"'s next to each other. if they swap places, does that count as a new instance? It is still the same thing really. That would involve another step. Whatta say?

In probability and counting the two N's are called indinstinguishable elements, so having them in two places and then swapping them counts as only one possibility, not two. That's why he is dividing by 2!3!3!...2 C's, 3 N's, and 3 I's.

 

[chuckywang]

Originally posted by: Evadman
Depends on how you determine what is being arranged. There are 2 "n"'s next to each other. if they swap places, does that count as a new instance? It is still the same thing really. That would involve another step. Whatta say?

No, they don't count as a new instance.

 

[TuxDave]

Originally posted by: Evadman
Depends on how you determine what is being arranged. There are 2 "n"'s next to each other. if they swap places, does that count as a new instance? It is still the same thing really. That would involve another step. Whatta say?

swapping the same letter doesn't count as a new instance based on the op's equation

 

[Random Variable]

I think it's [7!/(3!*2!)]*C(8,3)

First you have to permute cncnnat. Then the three i's can be placed where the asteriks are: *c*n*c*n*n*a*t*

 

[Random Variable]

I get 23,520 ways. Is that the answer?

 

[chuckywang]

Originally posted by: Vespasian
I think it's [7!/(3!*2!)]*C(8,3)

I think it's [7!/(3!*2!)]*8*8....maybe

EDIT: Whoops!! Geez, for a second I thought the OP wanted the number of ways 2 I's ARE adjacent. I agree with you. If you compute the stuff in my post above, you will find that our answers agree. But your way is much simpler. I should have thought of that sooner. Good job.

 

[TechnoKid]

i thnk its 42? yeah...that must be it...

 

[Injury]

Originally posted by: nakedfrog
I have a better question. Who gives a rat's arse?

WINNAR!