# A string is an ordering of letters ILLINOIS a

1. Aug 22, 2011

### look416

1. The problem statement, all variables and given/known data
In this problem a string is an ordering of letters ILLINOIS
a. how many such strings are there
b. in how many strings, such as LSLOINII, does each L appear before each I?
c. In how many strings does one I appear the L?

3. The attempt at a solution
a. this is pretty easy, just using partition formula
$\frac{8!}{2!3!1!1!1!}$
which is 3360

however, for question 2 and 3 im not so understand about it
for question 2, considering each L appear before each I, should i conclude them as a partition?
which eventually, leads to 8C5 $\times$ 3C1 $\times$ 2C1 $\times$ 1C1
need some hints for question 3
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 23, 2011

### Quinzio

Re: Combinatorics

For part b. I'd go like this:
Imagine for each string you highlight the Ls and Is.
Eg. ILLINOIS --> ILLII
How many possible dispositions you might get ?
They are $$\frac{5!}{2!3!}= 10$$
only one of them is the "good" one (LLIII) so, I'd say only 1 out of 10 of all the possible original combinations is good. That is 3360/10 = 336

Part c.
I assume the correct question is "In how many strings does one I appear before the L?"

The reasoning is pretty the same as before you just have one I stuck at the beginning
ILLII
Then you have to find in how many ways you can reaggange the four remaining letters: LLII
which is similarly as before $$\frac{4!}{2!2!}=6$$
So I'd say only 1 out of 6 of the original arrangements (560).

I just tried as you did, don't take my solutions for sure.

3. Aug 23, 2011

### look416

Re: Combinatorics

wow, you open another door for me to see about it, because typically, i still struggle in the ways to determining the correct ways to find the actual answer
thx for teaching me such a nice approach to these problems

Last edited: Aug 23, 2011
4. Aug 24, 2011

### Quinzio

Re: Combinatorics

In part c. there's a mistake:
it must be read 6 out of 10 of all possible original rearrangements:
3360*(6/10) = 2016

5. Aug 24, 2011

### look416

Re: Combinatorics

hmmm, why is it required to be assuming there must be 6 out of 10? i dun get the statement here

6. Aug 24, 2011

### Quinzio

Re: Combinatorics

Well, we can count them, there's not a lot of them.

L: 0
I: 1

1) 00111
2) 01011
3) 01101
4) 01110
5) 10011
6) 10101
7) 10110
8) 11001
9) 11010
10) 11100