Question 1154211
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If only the numbers of shells of the different types counts, there are 15 combinations:<br>
(4,0,0), (0,4,0), (0,0,4)
(3,1,0), (3,0,1), (1,3,0), (0,3,1), (1,0,3), (0,1,3)
(2,2,0), (2,0,2), (0,2,2)
(2,1,1), (1,2,1), (1,1,2)<br>
This is a common type of problem, adding a certain number of whole numbers to get a certain total.  This kind of problem can be solved using "stars and bars".<br>
Imagine picking up the shells and putting them in one of three bins to separate them by the type of shell.<br>
Now model that by imagining the four shells arranged in a row, and you place two dividers between the shells to represent dividing them into the three types.  Note that only (n-1) dividers are needed to separate the items into n types.<br>
Represent the shells with stars and the dividers with bars.  We have a string of six items -- the four shells and the two dividers.<br>
Each different arrangement of the six items represents a different distribution of the four shells among the three types. For example,<br>
**|*|*  represents two shells of the first type and one each of the other two types;
|***|*  represents three shells of the second type and one of the third.<br>
By a well-known counting principle, the number of different arrangements of those four stars and two bars is<br>
{{{6!/(4!)(2!) = C(6,2) = (6*5)/(2*1) = 30/2 = 15}}}<br>
In general, if there are n items to be divided into k groups, there are n stars and (k-1) bars, and the number of ways of distributing the items is<br>
{{{(n+k-1)!/((n!)(k-1)!)}}}<br>
Note a similar kind of problem requires that the total be composed of POSITIVE integers (i.e.,  there must be at least one item of each type).  In that case, simply take one item an put it in each bin; then use stars and bars to distribute the remaining (n-k) items.<br>
For this problem, we start with one of each of the three types of shells, leaving us with one more shell to "distribute" among the three types.  Obviously there are three ways to do that, as can be seen in the list of possible distributions at the beginning of this response.<br>
Using stars and bars, that is represented by the possible arrangements of the symbols<br>
*||<br>
There are obviously three different arrangements; if you wanted to use the formula to get that total, it would look like<br>
{{{(1+(3-1))!/((1!)(3-1)!) = 3!/((1!)(2!)) = 6/2 = 3}}}<br>