Question 1154211
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The space of events consists of all  4-digit numbers written with digits  1,  2  and  3,

where each digit represents one of the three types of shells.


Repeating is allowed for these digits.


The number of all elements  (of all 3-digit numbers of this kind)  is   {{{3^4}}} = 81,  because in each of the  4  positions any of  3  digits may occur.


The probability for each individual element of the space of events is    {{{1/81}}}.


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The solution above works when the order of shells in the collection is important (does matter).


We may consider other extreme case when the order of shells in the collection DOES NOT MATTER (is not important).


Then the set of events is the set of all different solutions to equation


                {{{x[1]}}} + {{{x[2]}}} + {{{x[3]}}} = 4        (1)


in integer non-negative numbers. 


Here each of the numbers  {{{x[1]}}},  {{{x[2]}}},  {{{x[3]}}} represents the  MULTIPLICITY  of the shell of the type  "i"  in the collection,   i = 1, 2, 3.


There is the formula for the number of solution to equation  (1)  under given restriction


            the number of solution = {{{((4+(3-1))!)/(4!*(3-1)!)}}} = {{{(4+2)!/(4!*2!)}}} = {{{6!/(4!*2!)}}} = {{{720/(24*2)}}} = {{{720/48}}} = 15.


Thus in this case the space of events has 15 elements with equal probability  of  {{{1/15}}}   each.


The method calculating the number of solutions to equation  (1) is called the  "Stars  and  Bars  method".


You can read about it in the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =https://www.algebra.com/algebra/homework/Permutations/Stars-and-bars-method-for-Combinatorics-problems-2.lesson>Stars and bars method for Combinatorics problems</A> 

in this site, &nbsp;&nbsp;Problem &nbsp;2.



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Notice after reading the post by @greenestamps.


<pre>
    The formula in the post by @greenestamps is correct, and his answer is correct, too.

    The formula in my original post was incorrect.

    Many thanks to greenestamps for noticing it (!)


    After detecting my error, I fixed my formula and re-calculated the answer.

    Now you see the edited version in my post, which agrees with the solution by @greenestamps.


    Again, many thanks to @greenestamps (!)


    It is nice to have so attentive, accurate and knowledgeable colleague !
</pre>

 
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<U>Comment from student</U>: Hi ikleyn, Thank you for your urgent response to my question. 
However, I am a bit confused with your answer. The reason is because, I looked at this task in the context of probability. 
If so, can we have a probability greater than 1?
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<U>My response</U>.


<pre>
Hello, I got this comment from a visitor. But since it came with no reference to the problem,
I don't know to WHICH PROBLEM it relates.


If it relates to THIS problem, then where do you see the word "probability" in your post?


In the future, if you want to have a DIALOG, then PLEASE REFER to the ID number of the problem, which is  1154211  in this case.


If the comment and the question do not relate to this problem, then ignore/disregard my post.


Have a nice day (!)
</pre>