A string is an ordering of letters ILLINOIS a

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The discussion revolves around calculating the number of distinct arrangements of the letters in "ILLINOIS." For part a, the total arrangements are calculated using the partition formula, resulting in 3360. In part b, the focus is on arrangements where each L appears before each I, leading to a conclusion of 336 valid combinations. Part c addresses the scenario where one I appears before the L, with the correct calculation yielding 2016 arrangements. The conversation highlights the complexities of combinatorial reasoning in string arrangements.
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Homework Statement


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?


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 I am 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
 
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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.
 


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


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


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
 
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